Traceback (most recent call last):
  File "/mnt/data3/yqlai/code/llama2-70B-LoRA/quantize/codellama/../quantize.py", line 2, in <module>
    from transformers import AutoModelForCausalLM, AutoTokenizer
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/__init__.py", line 26, in <module>
    from . import dependency_versions_check
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/dependency_versions_check.py", line 57, in <module>
    require_version_core(deps[pkg])
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 117, in require_version_core
    return require_version(requirement, hint)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 111, in require_version
    _compare_versions(op, got_ver, want_ver, requirement, pkg, hint)
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 44, in _compare_versions
    raise ImportError(
ImportError: tokenizers>=0.11.1,!=0.11.3,<0.14 is required for a normal functioning of this module, but found tokenizers==0.14.1.
Try: pip install transformers -U or pip install -e '.[dev]' if you're working with git main
clipping is on, will clip to 1000
2023-11-15 20:19:36.482 | INFO     | __main__:main:22 - loaded 24926 examples
2023-11-15 20:19:36.483 | INFO     | __main__:main:23 - shuffling 24926 examples
2023-11-15 20:19:36.515 | INFO     | __main__:main:25 - will use 2048 examples
The following is a conversation between human and assistant. 

Human: Find the angle between the vectors $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},$ in degrees.

Assistant: If $\theta$ is the angle between the vectors, then
\[\cos \theta = \frac{\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\| \left\| \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \right\|} = \frac{(2)(-1) + (-1)(1) + (1)(0)}{\sqrt{6} \cdot \sqrt{2}} = \frac{-3}{2 \sqrt{3}} = -\frac{\sqrt{3}}{2}.\]Hence, $\theta = 150^\circ.$</s>
=======================
The following is a conversation between human and assistant. 

Human: You find yourself in a magical forest filled with glowing trees, bioluminescent flowers, and mystical creatures. The air is filled with the sweet scent of blooming flowers and the soft sound of rustling leaves. In the distance, you see a majestic waterfall cascading into a crystal-clear pool.
Guide me through this enchanted forest.

Assistant: As we begin our journey, we walk along a moss-covered path that gently winds through the forest. You notice that the vibrant colors of the flora seem to pulse with energy, casting an ethereal glow on everything around us. We approach a group of friendly fairies who are playfully dancing around a cluster of luminescent mushrooms. They giggle and invite you to join them for a moment before continuing on your way.
Further down the path, we come across a small clearing where a family of unicorns is peacefully grazing. Their shimmering manes catch the light as they move gracefully through the tall grass. One of the younger unicorns approaches you curiously, allowing you to gently stroke its velvety muzzle before it trots back to its family.
We continue walking until we reach the base of the mesmerizing waterfall. As you gaze at the water cascading down from above, you can't help but feel a sense of serenity wash over you. The misty spray cools your face, and you decide to take off your shoes and dip your feet into the refreshing pool below.
As you sit there, enjoying the soothing sounds of the waterfall, a wise old owl perches on a nearby branch. It tilts its head and seems to be studying you intently. Sensing your curiosity, the owl shares some ancient wisdom about the forest and its enchanting inhabitants. With newfound knowledge and appreciation for this magical place, you thank the owl and prepare to leave the waterfall behind.
As we make our way back through the forest, the sun begins to set, casting a warm golden light through the trees. The glowing plants and creatures seem even more magical in the twilight, creating an unforgettable scene that will stay with you long after you leave this enchanted forest.</s>
=======================
The following is a conversation between human and assistant. 

Human: What do these two changes have in common?
water evaporating from a puddle
dry ice sublimating and becoming a gas
A: Both are caused by cooling.
B: Both are only physical changes.
C: Both are chemical changes.

Assistant: Step 1: Think about each change.
Water evaporating from a puddle is a change of state. So, it is a physical change. The liquid changes into a gas, but a different type of matter is not formed.
Dry ice is solid carbon dioxide. When dry ice gets warm, it changes state and becomes carbon dioxide gas. This change of state, from solid to gas, is called sublimation.
Dry ice becoming a gas is a physical change. A change of state does not form a different type of matter.
Step 2: Look at each answer choice.
Both are only physical changes.
Both changes are physical changes. No new matter is created.
Both are chemical changes.
Both changes are physical changes. They are not chemical changes.
Both are caused by heating.
Both changes are caused by heating.
Both are caused by cooling.
Neither change is caused by cooling.</s>
=======================
The following is a conversation between human and assistant. 

Human: In the circle below, $\overline{AB} \| \overline{CD}$. $\overline{AD}$ is a diameter of the circle, and $AD = 36^{\prime \prime}$. What is the number of inches in the length of $\widehat{AB}$? Express your answer in terms of $\pi$. [asy]
import markers;
import olympiad; size(150); import geometry; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,36));
pair A = 36*dir(130);
pair B = 36*dir(50);
pair D = 36*dir(-50);
pair C = 36*dir(-130);
draw(B--A--D--C);
label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$D$",D,E);
markangle("$50^{\circ}$",D,A,B,radius=10);
[/asy]

Assistant: By symmetry, $\widehat{BD}=\widehat{CA}=100^\circ$.  Furthermore, $\widehat{AB}=\widehat{CD}$, so \[360^\circ=\widehat{AB}+\widehat{BD}+\widehat{DC}+\widehat{CA}=2\widehat{AB}+200^\circ.\] Therefore the arc $\widehat{AB}$ measures $80^\circ$.  Since the diameter of the circle is $36''$, the length of the arc is \[\frac{80}{360}(\pi\cdot36)=8\pi}\text{~inches.\]</s>
=======================
The following is a conversation between human and assistant. 

Human: Given an integer `n`, return _a binary string representing its representation in base_ `-2`.
**Note** that the returned string should not have leading zeros unless the string is `"0 "`.
**Example 1:**
**Input:** n = 2
**Output:**  "110 "
**Explantion:** (-2)2 + (-2)1 = 2
**Example 2:**
**Input:** n = 3
**Output:**  "111 "
**Explantion:** (-2)2 + (-2)1 + (-2)0 = 3
**Example 3:**
**Input:** n = 4
**Output:**  "100 "
**Explantion:** (-2)2 = 4
**Constraints:**
*   `0 <= n <= 109`


Assistant: 
def odd_even_jumps(arr):
    n = len(arr)
    can_reach_odd = [False] * n
    can_reach_even = [False] * n
    can_reach_odd[-1] = can_reach_even[-1] = True
    index_map = {}
    index_map[arr[-1]] = n - 1
    good_indices = 1
    for i in range(n - 2, -1, -1):
        odd_jump = next(iter([value for value in index_map if value >= arr[i]]), None)
        even_jump = next(iter([value for value in index_map if value <= arr[i]]), None)
        if odd_jump is not None:
            can_reach_odd[i] = can_reach_even[index_map[odd_jump]]
        if even_jump is not None:
            can_reach_even[i] = can_reach_odd[index_map[even_jump]]
        if can_reach_odd[i]:
            good_indices += 1
        index_map[arr[i]] = i
    return good_indices
</s>
=======================
The following is a conversation between human and assistant. 

Human: During a science experiment at school, three students - Mia, Noah, and Sophie - each place a different colored liquid (red, blue, and green) into separate beakers. They then leave the room for lunch. While they're gone, another student accidentally knocks over the beakers, causing the liquids to mix together and create a purple solution. When Mia, Noah, and Sophie return, they don't know what happened. What do they think about the color of the liquid?

Assistant: Mia, Noah, and Sophie would likely be confused about the color of the liquid since it is now purple instead of their original individual colors (red, blue, and green). They might think that some sort of chemical reaction occurred between the liquids to create this new color.</s>
=======================
Special tokens have been added in the vocabulary, make sure the associated word embeddings are fine-tuned or trained.
The following is a conversation between human and assistant. 

Human: What is the value of $\frac{2013^3-2 \cdot 2013^2 \cdot 2014+3 \cdot 2013 \cdot 2014^2-2014^3+1}{2013 \cdot 2014}$?

Assistant: Let $a=2013$. The expression is equal to $\frac{a^3-2a^2(a+1)+3a(a+1)^2-(a+1)^3+1}{a(a+1)}$. We notice a common factor of $a(a+1)$ in the second and third terms of the numerator, so we split the fraction apart: $$\frac{-2a^2(a+1)+3a(a+1)^2}{a(a+1)}+\frac{a^3-(a+1)^3+1}{a(a+1)}$$The first part of the expression is equal to $-2a+3(a+1)$, which simplifies to $a+3$.
The second part of the expression has a numerator that can be factored as a difference of cubes. We will concentrate on the numerator for now: \begin{align*}
a^3-(a+1)^3+1 &= (a-(a+1))(a^2+a(a+1)+(a+1)^2)+1 \\
&= (-1)(a^2+a^2+a+a^2+2a+1)+1 \\
&= (-1)(3a^2+3a+1)+1 \\
&= -3(a^2+a)-1+1 \\
&= -3a(a+1)
\end{align*}Alternatively, we can expand the numerator, and factor:
\begin{align*}
a^3 - (a + 1)^3 + 1 &= a^3 - (a + 1)(a + 1)(a + 1) + 1 \\
&= a^3 - (a^2 + 2a + 1)(a + 1) + 1 \\
&= a^3 - (a^3 + 3a^2 + 3a + 1) + 1 \\
&= a^3 - a^3 - 3a^2 - 3a - 1 + 1 \\
&= -3a^2 - 3a \\
&= -3(a^2 + a) \\
&= -3a(a + 1).
\end{align*}Taken with the denominator, the second half of the expression simply equals $\frac{-3a(a+1)}{a(a+1)}=-3$.
Putting it back with the first half of the expression, the final expression equals $(a+3)+(-3)=a=2013$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Compute $\sin 120^\circ$.

Assistant: Let $P$ be the point on the unit circle that is $120^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(120)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,NW);
//label("$A$",A,SE);
label("$D$",D,S);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$.  Therefore, the coordinates of $P$ are $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, so $\sin 120^\circ = \frac{\sqrt{3}}{2}$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Find constants $A,$ $B,$ $C,$ and $D$ such that
\[\frac{4x^3 - 20x^2 + 37x  -25}{(x-2)^3(x-1)} = \frac{A}{x - 1} + \frac{B}{(x -2)^3} + \frac{C}{(x-2)^2}+\frac{D}{x-2}.\]Enter the order quadruple $(A,B,C,D).$

Assistant: Multiplying both sides by $(x-2)^3(x-1)$ gives us
$$4x^3 - 20x^2 + 37x  -25 = A(x -2)^3+B(x-1)+C(x -1)(x -2)+D(x-1)(x -2)^2.$$Setting $x=2$ gives $4(8)-20(4)+74-25=B$. Evaluating the expression on the left gives us $B=1$.
Setting $x=1$ gives $4-20+37-25=A(-1)^3$, and so $A=4$.
We still need to find $C$ and $D$. By choosing 2 new values for $x$, we can get two equations which we can solve for $C$ and $D$. We can pick convenient values to make our work easier.
When $x=0$, we get
$$-25=4(-2)^3+(-1)+C(-1)(-2)+D(-1)(-2)^2$$which simplifies to
$$2C-D=8.$$When $x=-1$, we get
$$4(-1)^3-20(-1)^2+37(-1)-25=4(-3)^3+(-2)+C(-2)(-3)+D(-2)(-3)^2$$which simplifies to
$$C-3D=4.$$We can multiply this equation by $2$ and subtract it from the previous one to get $-D+6D=8-2\cdot4=0$ and hence $D=0$. Then $2C=8$ and $C=4$.  Therefore $(A,B,C,D)=(4,1,4,0).$</s>
=======================
The following is a conversation between human and assistant. 

Human: Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be three vectors such that $\|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{c}\| = 2.$  Also, the angle between any two of these vectors is $\arccos \frac{5}{8}.$  Find the volume of the parallelepiped generated by $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}.$

Assistant: Let $\mathbf{p}$ be the projection of $\mathbf{c}$ onto the plane containing $\mathbf{a}$ and $\mathbf{b}.$
[asy]
import three;
size(140);
currentprojection = perspective(6,3,2);
real t = 60, k = Cos(t);
triple A, B, C, O, P, Q;
A = (Cos(t/2),Sin(t/2),0);
B = (Cos(t/2),-Sin(t/2),0);
C = (k/Cos(t/2),0,sqrt(1 - k^2/Cos(t/2)^2));
O = (0,0,0);
P = (k/Cos(t/2),0,0);
Q = k/(k + 1)*A + k/(k + 1)*B;
draw(O--A,Arrow3(6));
draw(O--B,Arrow3(6));
draw(O--C,Arrow3(6));
draw(O--P,Arrow3(6));
draw(C--P,dashed);
label("$\mathbf{a}$", A, S, fontsize(10));
label("$\mathbf{b}$", B, W, fontsize(10));
label("$\mathbf{c}$", C, NW, fontsize(10));
label("$\mathbf{p}$", P, SW, fontsize(10));
[/asy]
Then
\[\mathbf{p} = s \mathbf{a} + t \mathbf{b}\]for some scalars $s$ and $t.$  Let $\mathbf{n}$ be the normal vector to the plane containing $\mathbf{a}$ and $\mathbf{b},$ so
\[\mathbf{c} = \mathbf{p} + u \mathbf{n} = s \mathbf{a} + t \mathbf{b} + u \mathbf{n}\]for some scalar $u.$
Taking the dot product with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{a} + t \mathbf{a} \cdot \mathbf{b} + u \mathbf{a} \cdot \mathbf{n}.\]Note that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 4$ and $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 2 \cdot 2 \cdot \frac{5}{8} = \frac{5}{2}.$  Also, $\mathbf{a} \cdot \mathbf{n} = 0,$ so
\[\frac{5}{2} = 4s + \frac{5t}{2}.\]Similarly, taking the dot product with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{b} + t \mathbf{b} \cdot \mathbf{b} + u \mathbf{b} \cdot \mathbf{n}.\]This reduces to $\frac{5}{2} = \frac{5s}{2} + 4t.$
Solving the equations $\frac{5}{2} = 4s + \frac{5t}{2}$ and $\frac{5}{2} = \frac{5s}{2} + 4t,$ we get $s = t = \frac{5}{13}.$  Hence,
\[\mathbf{p} = \frac{5}{13} (\mathbf{a} + \mathbf{b}).\]Then
\begin{align*}
\|\mathbf{p}\|^2 &= \frac{25}{169} (\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) \\
&= \frac{25}{169} \left( 4 + 2 \cdot \frac{5}{2} + 4 \right) = \frac{25}{13}.
\end{align*}By Pythagoras, the height of the parallelepiped is then given by
\[\sqrt{4 - \|\mathbf{p}\|^2} = \sqrt{4 - \frac{25}{13}} = \sqrt{\frac{27}{13}}.\]The base of the parallelepiped has area $2 \cdot 2 \cdot \sin \left( \arccos \frac{5}{8} \right) = 4 \sqrt{1 - \left( \frac{5}{8} \right)^2} = 4 \sqrt{\frac{39}{64}},$ so the volume of the parallelepiped is
\[\sqrt{\frac{27}{13}} \cdot 4 \sqrt{\frac{39}{64}} = \frac{9}{2}.\]</s>
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Traceback (most recent call last):
  File "/mnt/data3/yqlai/code/llama2-70B-LoRA/quantize/codellama/../quantize.py", line 123, in <module>
    main(args)
  File "/mnt/data3/yqlai/code/llama2-70B-LoRA/quantize/codellama/../quantize.py", line 66, in main
    quantized_model = quantizer.quantize_model(model, tokenizer)
  File "/mnt/data3/anaconda3/envs/llama/lib/python3.9/site-packages/torch/utils/_contextlib.py", line 115, in decorate_context
    return func(*args, **kwargs)
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/optimum/optimum/gptq/quantizer.py", line 419, in quantize_model
    scale, zero, g_idx = gptq[name].fasterquant(
  File "/mnt/data3/anaconda3/envs/llama/lib/python3.9/site-packages/auto_gptq/quantization/gptq.py", line 111, in fasterquant
    H = torch.cholesky_inverse(H)
torch.cuda.OutOfMemoryError: CUDA out of memory. Tried to allocate 1.81 GiB (GPU 0; 39.45 GiB total capacity; 21.33 GiB already allocated; 1.75 GiB free; 36.41 GiB reserved in total by PyTorch) If reserved memory is >> allocated memory try setting max_split_size_mb to avoid fragmentation.  See documentation for Memory Management and PYTORCH_CUDA_ALLOC_CONF
clipping is on, will clip to 1000
2023-11-15 21:07:41.083 | INFO     | __main__:main:22 - loaded 24926 examples
2023-11-15 21:07:41.084 | INFO     | __main__:main:23 - shuffling 24926 examples
2023-11-15 21:07:41.100 | INFO     | __main__:main:25 - will use 1024 examples
The following is a conversation between human and assistant. 

Human: Find the angle between the vectors $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},$ in degrees.

Assistant: If $\theta$ is the angle between the vectors, then
\[\cos \theta = \frac{\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\| \left\| \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \right\|} = \frac{(2)(-1) + (-1)(1) + (1)(0)}{\sqrt{6} \cdot \sqrt{2}} = \frac{-3}{2 \sqrt{3}} = -\frac{\sqrt{3}}{2}.\]Hence, $\theta = 150^\circ.$</s>
=======================
The following is a conversation between human and assistant. 

Human: You find yourself in a magical forest filled with glowing trees, bioluminescent flowers, and mystical creatures. The air is filled with the sweet scent of blooming flowers and the soft sound of rustling leaves. In the distance, you see a majestic waterfall cascading into a crystal-clear pool.
Guide me through this enchanted forest.

Assistant: As we begin our journey, we walk along a moss-covered path that gently winds through the forest. You notice that the vibrant colors of the flora seem to pulse with energy, casting an ethereal glow on everything around us. We approach a group of friendly fairies who are playfully dancing around a cluster of luminescent mushrooms. They giggle and invite you to join them for a moment before continuing on your way.
Further down the path, we come across a small clearing where a family of unicorns is peacefully grazing. Their shimmering manes catch the light as they move gracefully through the tall grass. One of the younger unicorns approaches you curiously, allowing you to gently stroke its velvety muzzle before it trots back to its family.
We continue walking until we reach the base of the mesmerizing waterfall. As you gaze at the water cascading down from above, you can't help but feel a sense of serenity wash over you. The misty spray cools your face, and you decide to take off your shoes and dip your feet into the refreshing pool below.
As you sit there, enjoying the soothing sounds of the waterfall, a wise old owl perches on a nearby branch. It tilts its head and seems to be studying you intently. Sensing your curiosity, the owl shares some ancient wisdom about the forest and its enchanting inhabitants. With newfound knowledge and appreciation for this magical place, you thank the owl and prepare to leave the waterfall behind.
As we make our way back through the forest, the sun begins to set, casting a warm golden light through the trees. The glowing plants and creatures seem even more magical in the twilight, creating an unforgettable scene that will stay with you long after you leave this enchanted forest.</s>
=======================
The following is a conversation between human and assistant. 

Human: What do these two changes have in common?
water evaporating from a puddle
dry ice sublimating and becoming a gas
A: Both are caused by cooling.
B: Both are only physical changes.
C: Both are chemical changes.

Assistant: Step 1: Think about each change.
Water evaporating from a puddle is a change of state. So, it is a physical change. The liquid changes into a gas, but a different type of matter is not formed.
Dry ice is solid carbon dioxide. When dry ice gets warm, it changes state and becomes carbon dioxide gas. This change of state, from solid to gas, is called sublimation.
Dry ice becoming a gas is a physical change. A change of state does not form a different type of matter.
Step 2: Look at each answer choice.
Both are only physical changes.
Both changes are physical changes. No new matter is created.
Both are chemical changes.
Both changes are physical changes. They are not chemical changes.
Both are caused by heating.
Both changes are caused by heating.
Both are caused by cooling.
Neither change is caused by cooling.</s>
=======================
The following is a conversation between human and assistant. 

Human: In the circle below, $\overline{AB} \| \overline{CD}$. $\overline{AD}$ is a diameter of the circle, and $AD = 36^{\prime \prime}$. What is the number of inches in the length of $\widehat{AB}$? Express your answer in terms of $\pi$. [asy]
import markers;
import olympiad; size(150); import geometry; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,36));
pair A = 36*dir(130);
pair B = 36*dir(50);
pair D = 36*dir(-50);
pair C = 36*dir(-130);
draw(B--A--D--C);
label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$D$",D,E);
markangle("$50^{\circ}$",D,A,B,radius=10);
[/asy]

Assistant: By symmetry, $\widehat{BD}=\widehat{CA}=100^\circ$.  Furthermore, $\widehat{AB}=\widehat{CD}$, so \[360^\circ=\widehat{AB}+\widehat{BD}+\widehat{DC}+\widehat{CA}=2\widehat{AB}+200^\circ.\] Therefore the arc $\widehat{AB}$ measures $80^\circ$.  Since the diameter of the circle is $36''$, the length of the arc is \[\frac{80}{360}(\pi\cdot36)=8\pi}\text{~inches.\]</s>
=======================
The following is a conversation between human and assistant. 

Human: Given an integer `n`, return _a binary string representing its representation in base_ `-2`.
**Note** that the returned string should not have leading zeros unless the string is `"0 "`.
**Example 1:**
**Input:** n = 2
**Output:**  "110 "
**Explantion:** (-2)2 + (-2)1 = 2
**Example 2:**
**Input:** n = 3
**Output:**  "111 "
**Explantion:** (-2)2 + (-2)1 + (-2)0 = 3
**Example 3:**
**Input:** n = 4
**Output:**  "100 "
**Explantion:** (-2)2 = 4
**Constraints:**
*   `0 <= n <= 109`


Assistant: 
def odd_even_jumps(arr):
    n = len(arr)
    can_reach_odd = [False] * n
    can_reach_even = [False] * n
    can_reach_odd[-1] = can_reach_even[-1] = True
    index_map = {}
    index_map[arr[-1]] = n - 1
    good_indices = 1
    for i in range(n - 2, -1, -1):
        odd_jump = next(iter([value for value in index_map if value >= arr[i]]), None)
        even_jump = next(iter([value for value in index_map if value <= arr[i]]), None)
        if odd_jump is not None:
            can_reach_odd[i] = can_reach_even[index_map[odd_jump]]
        if even_jump is not None:
            can_reach_even[i] = can_reach_odd[index_map[even_jump]]
        if can_reach_odd[i]:
            good_indices += 1
        index_map[arr[i]] = i
    return good_indices
</s>
=======================
The following is a conversation between human and assistant. 

Human: During a science experiment at school, three students - Mia, Noah, and Sophie - each place a different colored liquid (red, blue, and green) into separate beakers. They then leave the room for lunch. While they're gone, another student accidentally knocks over the beakers, causing the liquids to mix together and create a purple solution. When Mia, Noah, and Sophie return, they don't know what happened. What do they think about the color of the liquid?

Assistant: Mia, Noah, and Sophie would likely be confused about the color of the liquid since it is now purple instead of their original individual colors (red, blue, and green). They might think that some sort of chemical reaction occurred between the liquids to create this new color.</s>
=======================
Special tokens have been added in the vocabulary, make sure the associated word embeddings are fine-tuned or trained.
The following is a conversation between human and assistant. 

Human: What is the value of $\frac{2013^3-2 \cdot 2013^2 \cdot 2014+3 \cdot 2013 \cdot 2014^2-2014^3+1}{2013 \cdot 2014}$?

Assistant: Let $a=2013$. The expression is equal to $\frac{a^3-2a^2(a+1)+3a(a+1)^2-(a+1)^3+1}{a(a+1)}$. We notice a common factor of $a(a+1)$ in the second and third terms of the numerator, so we split the fraction apart: $$\frac{-2a^2(a+1)+3a(a+1)^2}{a(a+1)}+\frac{a^3-(a+1)^3+1}{a(a+1)}$$The first part of the expression is equal to $-2a+3(a+1)$, which simplifies to $a+3$.
The second part of the expression has a numerator that can be factored as a difference of cubes. We will concentrate on the numerator for now: \begin{align*}
a^3-(a+1)^3+1 &= (a-(a+1))(a^2+a(a+1)+(a+1)^2)+1 \\
&= (-1)(a^2+a^2+a+a^2+2a+1)+1 \\
&= (-1)(3a^2+3a+1)+1 \\
&= -3(a^2+a)-1+1 \\
&= -3a(a+1)
\end{align*}Alternatively, we can expand the numerator, and factor:
\begin{align*}
a^3 - (a + 1)^3 + 1 &= a^3 - (a + 1)(a + 1)(a + 1) + 1 \\
&= a^3 - (a^2 + 2a + 1)(a + 1) + 1 \\
&= a^3 - (a^3 + 3a^2 + 3a + 1) + 1 \\
&= a^3 - a^3 - 3a^2 - 3a - 1 + 1 \\
&= -3a^2 - 3a \\
&= -3(a^2 + a) \\
&= -3a(a + 1).
\end{align*}Taken with the denominator, the second half of the expression simply equals $\frac{-3a(a+1)}{a(a+1)}=-3$.
Putting it back with the first half of the expression, the final expression equals $(a+3)+(-3)=a=2013$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Compute $\sin 120^\circ$.

Assistant: Let $P$ be the point on the unit circle that is $120^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(120)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,NW);
//label("$A$",A,SE);
label("$D$",D,S);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$.  Therefore, the coordinates of $P$ are $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, so $\sin 120^\circ = \frac{\sqrt{3}}{2}$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Find constants $A,$ $B,$ $C,$ and $D$ such that
\[\frac{4x^3 - 20x^2 + 37x  -25}{(x-2)^3(x-1)} = \frac{A}{x - 1} + \frac{B}{(x -2)^3} + \frac{C}{(x-2)^2}+\frac{D}{x-2}.\]Enter the order quadruple $(A,B,C,D).$

Assistant: Multiplying both sides by $(x-2)^3(x-1)$ gives us
$$4x^3 - 20x^2 + 37x  -25 = A(x -2)^3+B(x-1)+C(x -1)(x -2)+D(x-1)(x -2)^2.$$Setting $x=2$ gives $4(8)-20(4)+74-25=B$. Evaluating the expression on the left gives us $B=1$.
Setting $x=1$ gives $4-20+37-25=A(-1)^3$, and so $A=4$.
We still need to find $C$ and $D$. By choosing 2 new values for $x$, we can get two equations which we can solve for $C$ and $D$. We can pick convenient values to make our work easier.
When $x=0$, we get
$$-25=4(-2)^3+(-1)+C(-1)(-2)+D(-1)(-2)^2$$which simplifies to
$$2C-D=8.$$When $x=-1$, we get
$$4(-1)^3-20(-1)^2+37(-1)-25=4(-3)^3+(-2)+C(-2)(-3)+D(-2)(-3)^2$$which simplifies to
$$C-3D=4.$$We can multiply this equation by $2$ and subtract it from the previous one to get $-D+6D=8-2\cdot4=0$ and hence $D=0$. Then $2C=8$ and $C=4$.  Therefore $(A,B,C,D)=(4,1,4,0).$</s>
=======================
The following is a conversation between human and assistant. 

Human: Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be three vectors such that $\|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{c}\| = 2.$  Also, the angle between any two of these vectors is $\arccos \frac{5}{8}.$  Find the volume of the parallelepiped generated by $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}.$

Assistant: Let $\mathbf{p}$ be the projection of $\mathbf{c}$ onto the plane containing $\mathbf{a}$ and $\mathbf{b}.$
[asy]
import three;
size(140);
currentprojection = perspective(6,3,2);
real t = 60, k = Cos(t);
triple A, B, C, O, P, Q;
A = (Cos(t/2),Sin(t/2),0);
B = (Cos(t/2),-Sin(t/2),0);
C = (k/Cos(t/2),0,sqrt(1 - k^2/Cos(t/2)^2));
O = (0,0,0);
P = (k/Cos(t/2),0,0);
Q = k/(k + 1)*A + k/(k + 1)*B;
draw(O--A,Arrow3(6));
draw(O--B,Arrow3(6));
draw(O--C,Arrow3(6));
draw(O--P,Arrow3(6));
draw(C--P,dashed);
label("$\mathbf{a}$", A, S, fontsize(10));
label("$\mathbf{b}$", B, W, fontsize(10));
label("$\mathbf{c}$", C, NW, fontsize(10));
label("$\mathbf{p}$", P, SW, fontsize(10));
[/asy]
Then
\[\mathbf{p} = s \mathbf{a} + t \mathbf{b}\]for some scalars $s$ and $t.$  Let $\mathbf{n}$ be the normal vector to the plane containing $\mathbf{a}$ and $\mathbf{b},$ so
\[\mathbf{c} = \mathbf{p} + u \mathbf{n} = s \mathbf{a} + t \mathbf{b} + u \mathbf{n}\]for some scalar $u.$
Taking the dot product with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{a} + t \mathbf{a} \cdot \mathbf{b} + u \mathbf{a} \cdot \mathbf{n}.\]Note that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 4$ and $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 2 \cdot 2 \cdot \frac{5}{8} = \frac{5}{2}.$  Also, $\mathbf{a} \cdot \mathbf{n} = 0,$ so
\[\frac{5}{2} = 4s + \frac{5t}{2}.\]Similarly, taking the dot product with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{b} + t \mathbf{b} \cdot \mathbf{b} + u \mathbf{b} \cdot \mathbf{n}.\]This reduces to $\frac{5}{2} = \frac{5s}{2} + 4t.$
Solving the equations $\frac{5}{2} = 4s + \frac{5t}{2}$ and $\frac{5}{2} = \frac{5s}{2} + 4t,$ we get $s = t = \frac{5}{13}.$  Hence,
\[\mathbf{p} = \frac{5}{13} (\mathbf{a} + \mathbf{b}).\]Then
\begin{align*}
\|\mathbf{p}\|^2 &= \frac{25}{169} (\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) \\
&= \frac{25}{169} \left( 4 + 2 \cdot \frac{5}{2} + 4 \right) = \frac{25}{13}.
\end{align*}By Pythagoras, the height of the parallelepiped is then given by
\[\sqrt{4 - \|\mathbf{p}\|^2} = \sqrt{4 - \frac{25}{13}} = \sqrt{\frac{27}{13}}.\]The base of the parallelepiped has area $2 \cdot 2 \cdot \sin \left( \arccos \frac{5}{8} \right) = 4 \sqrt{1 - \left( \frac{5}{8} \right)^2} = 4 \sqrt{\frac{39}{64}},$ so the volume of the parallelepiped is
\[\sqrt{\frac{27}{13}} \cdot 4 \sqrt{\frac{39}{64}} = \frac{9}{2}.\]</s>
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Quantizing layers inside the block: 100%|██████████| 7/7 [02:24<00:00, 27.81s/it]

                                                                                 
Quantizing model.layers blocks :  10%|█         | 5/48 [13:15<1:53:13, 157.99s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:30<01:16, 15.25s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:31<00:15, 15.30s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:25<00:00, 27.89s/it]

                                                                                 
Quantizing model.layers blocks :  12%|█▎        | 6/48 [15:56<1:51:19, 159.03s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.95s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:15, 15.03s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:23<00:00, 27.52s/it]

                                                                                 
Quantizing model.layers blocks :  15%|█▍        | 7/48 [18:33<1:48:22, 158.60s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.99s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:59, 14.89s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:30<00:15, 15.05s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:23<00:00, 27.57s/it]

                                                                                 
Quantizing model.layers blocks :  17%|█▋        | 8/48 [21:08<1:44:50, 157.25s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.75s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:58, 14.70s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:14<00:30, 15.07s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:15, 15.05s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:24<00:00, 28.07s/it]

                                                                                 
Quantizing model.layers blocks :  19%|█▉        | 9/48 [23:46<1:42:27, 157.64s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.95s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:30<00:15, 15.20s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:24<00:00, 27.89s/it]

                                                                                 
Quantizing model.layers blocks :  21%|██        | 10/48 [26:25<1:40:02, 157.97s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:30<01:15, 15.09s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:45<01:00, 15.10s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:16<00:30, 15.26s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:31<00:15, 15.34s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:26<00:00, 28.27s/it]

                                                                                 
Quantizing model.layers blocks :  23%|██▎       | 11/48 [29:09<1:38:29, 159.72s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:30<01:15, 15.18s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:45<01:00, 15.16s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:30<00:14, 14.98s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:24<00:00, 27.71s/it]

                                                                                 
Quantizing model.layers blocks :  25%|██▌       | 12/48 [31:52<1:36:26, 160.74s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.94s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:45<01:00, 15.08s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:00<00:45, 15.17s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:15<00:30, 15.18s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:30<00:15, 15.07s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:23<00:00, 27.62s/it]

                                                                                 
Quantizing model.layers blocks :  27%|██▋       | 13/48 [34:30<1:33:15, 159.88s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:30<01:15, 15.19s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:45<00:59, 14.96s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:15<00:30, 15.06s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:30<00:15, 15.07s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:24<00:00, 27.89s/it]

                                                                                 
Quantizing model.layers blocks :  29%|██▉       | 14/48 [37:06<1:29:55, 158.71s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.77s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:14<00:30, 15.03s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:15, 15.08s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:23<00:00, 27.57s/it]

                                                                                 
Quantizing model.layers blocks :  31%|███▏      | 15/48 [39:40<1:26:29, 157.27s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.72s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:58, 14.73s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:14<00:29, 14.85s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.94s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.36s/it]

                                                                                 
Quantizing model.layers blocks :  33%|███▎      | 16/48 [42:12<1:23:10, 155.95s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.84s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:59, 14.86s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:14<00:29, 14.89s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.91s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.28s/it]

                                                                                 
Quantizing model.layers blocks :  35%|███▌      | 17/48 [44:46<1:20:14, 155.32s/it]

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Quantizing layers inside the block:  14%|█▍        | 1/7 [00:14<01:28, 14.82s/it]

Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.77s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:59, 14.76s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [00:58<00:44, 14.73s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:13<00:29, 14.79s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:28<00:14, 14.83s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.41s/it]

                                                                                 
Quantizing model.layers blocks :  38%|███▊      | 18/48 [47:19<1:17:15, 154.53s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.82s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:44<00:59, 14.87s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:14<00:29, 14.89s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.98s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.48s/it]

                                                                                 
Quantizing model.layers blocks :  40%|███▉      | 19/48 [49:53<1:14:34, 154.30s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.71s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.98s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.38s/it]

                                                                                 
Quantizing model.layers blocks :  42%|████▏     | 20/48 [52:26<1:11:50, 153.95s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.88s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.93s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.51s/it]

                                                                                 
Quantizing model.layers blocks :  44%|████▍     | 21/48 [54:59<1:09:13, 153.85s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.82s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.90s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:21<00:00, 27.30s/it]

                                                                                 
Quantizing model.layers blocks :  46%|████▌     | 22/48 [57:32<1:06:31, 153.53s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:14, 14.88s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.97s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.34s/it]

                                                                                 
Quantizing model.layers blocks :  48%|████▊     | 23/48 [1:00:08<1:04:16, 154.27s/it]

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Quantizing layers inside the block:  14%|█▍        | 1/7 [00:14<01:29, 14.86s/it]

Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.79s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:28<00:14, 14.84s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:21<00:00, 27.34s/it]

                                                                                 
Quantizing model.layers blocks :  50%|█████     | 24/48 [1:02:41<1:01:28, 153.70s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:29<01:13, 14.75s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:28<00:14, 14.83s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:21<00:00, 27.24s/it]

                                                                                 
Quantizing model.layers blocks :  52%|█████▏    | 25/48 [1:05:13<58:44, 153.23s/it]  

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.94s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.35s/it]

                                                                                 
Quantizing model.layers blocks :  54%|█████▍    | 26/48 [1:07:45<56:06, 153.01s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:29<00:14, 14.92s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.50s/it]

                                                                                 
Quantizing model.layers blocks :  56%|█████▋    | 27/48 [1:10:18<53:34, 153.07s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:28<00:14, 14.85s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:22<00:00, 27.34s/it]

                                                                                 
Quantizing model.layers blocks :  58%|█████▊    | 28/48 [1:12:51<50:57, 152.87s/it]

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Quantizing model.layers blocks : 100%|██████████| 48/48 [2:04:57<00:00, 157.49s/it]
Quantizing model.layers blocks : 100%|██████████| 48/48 [2:04:57<00:00, 156.21s/it]
Found modules on cpu/disk. Using Exllama backend requires all the modules to be on GPU. Setting `disable_exllama=True`
2023-11-16 00:36:36.273 | INFO     | __main__:main:67 - Quantizing model...done in 3.4404292696272023 hours
2023-11-16 00:36:36.285 | INFO     | __main__:main:68 - Saving model
2023-11-16 00:38:00.261 | INFO     | __main__:main:72 - done
Traceback (most recent call last):
  File "/mnt/data3/yqlai/code/llama2-70B-LoRA/quantize/codellama/../quantize.py", line 2, in <module>
    from transformers import AutoModelForCausalLM, AutoTokenizer
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/__init__.py", line 26, in <module>
    from . import dependency_versions_check
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/dependency_versions_check.py", line 57, in <module>
    require_version_core(deps[pkg])
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 117, in require_version_core
    return require_version(requirement, hint)
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 111, in require_version
    _compare_versions(op, got_ver, want_ver, requirement, pkg, hint)
  File "/mnt/data3/yqlai/code/nips2023_final/submodules/transformers/src/transformers/utils/versions.py", line 44, in _compare_versions
    raise ImportError(
ImportError: tokenizers>=0.11.1,!=0.11.3,<0.14 is required for a normal functioning of this module, but found tokenizers==0.14.1.
Try: pip install transformers -U or pip install -e '.[dev]' if you're working with git main
clipping is on, will clip to 1000
2023-11-16 00:50:42.944 | INFO     | __main__:main:22 - loaded 24926 examples
2023-11-16 00:50:42.945 | INFO     | __main__:main:23 - shuffling 24926 examples
2023-11-16 00:50:42.958 | INFO     | __main__:main:25 - will use 1280 examples
The following is a conversation between human and assistant. 

Human: Find the angle between the vectors $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},$ in degrees.

Assistant: If $\theta$ is the angle between the vectors, then
\[\cos \theta = \frac{\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\| \left\| \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \right\|} = \frac{(2)(-1) + (-1)(1) + (1)(0)}{\sqrt{6} \cdot \sqrt{2}} = \frac{-3}{2 \sqrt{3}} = -\frac{\sqrt{3}}{2}.\]Hence, $\theta = 150^\circ.$</s>
=======================
The following is a conversation between human and assistant. 

Human: You find yourself in a magical forest filled with glowing trees, bioluminescent flowers, and mystical creatures. The air is filled with the sweet scent of blooming flowers and the soft sound of rustling leaves. In the distance, you see a majestic waterfall cascading into a crystal-clear pool.
Guide me through this enchanted forest.

Assistant: As we begin our journey, we walk along a moss-covered path that gently winds through the forest. You notice that the vibrant colors of the flora seem to pulse with energy, casting an ethereal glow on everything around us. We approach a group of friendly fairies who are playfully dancing around a cluster of luminescent mushrooms. They giggle and invite you to join them for a moment before continuing on your way.
Further down the path, we come across a small clearing where a family of unicorns is peacefully grazing. Their shimmering manes catch the light as they move gracefully through the tall grass. One of the younger unicorns approaches you curiously, allowing you to gently stroke its velvety muzzle before it trots back to its family.
We continue walking until we reach the base of the mesmerizing waterfall. As you gaze at the water cascading down from above, you can't help but feel a sense of serenity wash over you. The misty spray cools your face, and you decide to take off your shoes and dip your feet into the refreshing pool below.
As you sit there, enjoying the soothing sounds of the waterfall, a wise old owl perches on a nearby branch. It tilts its head and seems to be studying you intently. Sensing your curiosity, the owl shares some ancient wisdom about the forest and its enchanting inhabitants. With newfound knowledge and appreciation for this magical place, you thank the owl and prepare to leave the waterfall behind.
As we make our way back through the forest, the sun begins to set, casting a warm golden light through the trees. The glowing plants and creatures seem even more magical in the twilight, creating an unforgettable scene that will stay with you long after you leave this enchanted forest.</s>
=======================
The following is a conversation between human and assistant. 

Human: What do these two changes have in common?
water evaporating from a puddle
dry ice sublimating and becoming a gas
A: Both are caused by cooling.
B: Both are only physical changes.
C: Both are chemical changes.

Assistant: Step 1: Think about each change.
Water evaporating from a puddle is a change of state. So, it is a physical change. The liquid changes into a gas, but a different type of matter is not formed.
Dry ice is solid carbon dioxide. When dry ice gets warm, it changes state and becomes carbon dioxide gas. This change of state, from solid to gas, is called sublimation.
Dry ice becoming a gas is a physical change. A change of state does not form a different type of matter.
Step 2: Look at each answer choice.
Both are only physical changes.
Both changes are physical changes. No new matter is created.
Both are chemical changes.
Both changes are physical changes. They are not chemical changes.
Both are caused by heating.
Both changes are caused by heating.
Both are caused by cooling.
Neither change is caused by cooling.</s>
=======================
The following is a conversation between human and assistant. 

Human: In the circle below, $\overline{AB} \| \overline{CD}$. $\overline{AD}$ is a diameter of the circle, and $AD = 36^{\prime \prime}$. What is the number of inches in the length of $\widehat{AB}$? Express your answer in terms of $\pi$. [asy]
import markers;
import olympiad; size(150); import geometry; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,36));
pair A = 36*dir(130);
pair B = 36*dir(50);
pair D = 36*dir(-50);
pair C = 36*dir(-130);
draw(B--A--D--C);
label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$D$",D,E);
markangle("$50^{\circ}$",D,A,B,radius=10);
[/asy]

Assistant: By symmetry, $\widehat{BD}=\widehat{CA}=100^\circ$.  Furthermore, $\widehat{AB}=\widehat{CD}$, so \[360^\circ=\widehat{AB}+\widehat{BD}+\widehat{DC}+\widehat{CA}=2\widehat{AB}+200^\circ.\] Therefore the arc $\widehat{AB}$ measures $80^\circ$.  Since the diameter of the circle is $36''$, the length of the arc is \[\frac{80}{360}(\pi\cdot36)=8\pi}\text{~inches.\]</s>
=======================
The following is a conversation between human and assistant. 

Human: Given an integer `n`, return _a binary string representing its representation in base_ `-2`.
**Note** that the returned string should not have leading zeros unless the string is `"0 "`.
**Example 1:**
**Input:** n = 2
**Output:**  "110 "
**Explantion:** (-2)2 + (-2)1 = 2
**Example 2:**
**Input:** n = 3
**Output:**  "111 "
**Explantion:** (-2)2 + (-2)1 + (-2)0 = 3
**Example 3:**
**Input:** n = 4
**Output:**  "100 "
**Explantion:** (-2)2 = 4
**Constraints:**
*   `0 <= n <= 109`


Assistant: 
def odd_even_jumps(arr):
    n = len(arr)
    can_reach_odd = [False] * n
    can_reach_even = [False] * n
    can_reach_odd[-1] = can_reach_even[-1] = True
    index_map = {}
    index_map[arr[-1]] = n - 1
    good_indices = 1
    for i in range(n - 2, -1, -1):
        odd_jump = next(iter([value for value in index_map if value >= arr[i]]), None)
        even_jump = next(iter([value for value in index_map if value <= arr[i]]), None)
        if odd_jump is not None:
            can_reach_odd[i] = can_reach_even[index_map[odd_jump]]
        if even_jump is not None:
            can_reach_even[i] = can_reach_odd[index_map[even_jump]]
        if can_reach_odd[i]:
            good_indices += 1
        index_map[arr[i]] = i
    return good_indices
</s>
=======================
The following is a conversation between human and assistant. 

Human: During a science experiment at school, three students - Mia, Noah, and Sophie - each place a different colored liquid (red, blue, and green) into separate beakers. They then leave the room for lunch. While they're gone, another student accidentally knocks over the beakers, causing the liquids to mix together and create a purple solution. When Mia, Noah, and Sophie return, they don't know what happened. What do they think about the color of the liquid?

Assistant: Mia, Noah, and Sophie would likely be confused about the color of the liquid since it is now purple instead of their original individual colors (red, blue, and green). They might think that some sort of chemical reaction occurred between the liquids to create this new color.</s>
=======================
Special tokens have been added in the vocabulary, make sure the associated word embeddings are fine-tuned or trained.
The following is a conversation between human and assistant. 

Human: What is the value of $\frac{2013^3-2 \cdot 2013^2 \cdot 2014+3 \cdot 2013 \cdot 2014^2-2014^3+1}{2013 \cdot 2014}$?

Assistant: Let $a=2013$. The expression is equal to $\frac{a^3-2a^2(a+1)+3a(a+1)^2-(a+1)^3+1}{a(a+1)}$. We notice a common factor of $a(a+1)$ in the second and third terms of the numerator, so we split the fraction apart: $$\frac{-2a^2(a+1)+3a(a+1)^2}{a(a+1)}+\frac{a^3-(a+1)^3+1}{a(a+1)}$$The first part of the expression is equal to $-2a+3(a+1)$, which simplifies to $a+3$.
The second part of the expression has a numerator that can be factored as a difference of cubes. We will concentrate on the numerator for now: \begin{align*}
a^3-(a+1)^3+1 &= (a-(a+1))(a^2+a(a+1)+(a+1)^2)+1 \\
&= (-1)(a^2+a^2+a+a^2+2a+1)+1 \\
&= (-1)(3a^2+3a+1)+1 \\
&= -3(a^2+a)-1+1 \\
&= -3a(a+1)
\end{align*}Alternatively, we can expand the numerator, and factor:
\begin{align*}
a^3 - (a + 1)^3 + 1 &= a^3 - (a + 1)(a + 1)(a + 1) + 1 \\
&= a^3 - (a^2 + 2a + 1)(a + 1) + 1 \\
&= a^3 - (a^3 + 3a^2 + 3a + 1) + 1 \\
&= a^3 - a^3 - 3a^2 - 3a - 1 + 1 \\
&= -3a^2 - 3a \\
&= -3(a^2 + a) \\
&= -3a(a + 1).
\end{align*}Taken with the denominator, the second half of the expression simply equals $\frac{-3a(a+1)}{a(a+1)}=-3$.
Putting it back with the first half of the expression, the final expression equals $(a+3)+(-3)=a=2013$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Compute $\sin 120^\circ$.

Assistant: Let $P$ be the point on the unit circle that is $120^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(120)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,NW);
//label("$A$",A,SE);
label("$D$",D,S);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$.  Therefore, the coordinates of $P$ are $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, so $\sin 120^\circ = \frac{\sqrt{3}}{2}$.</s>
=======================
The following is a conversation between human and assistant. 

Human: Find constants $A,$ $B,$ $C,$ and $D$ such that
\[\frac{4x^3 - 20x^2 + 37x  -25}{(x-2)^3(x-1)} = \frac{A}{x - 1} + \frac{B}{(x -2)^3} + \frac{C}{(x-2)^2}+\frac{D}{x-2}.\]Enter the order quadruple $(A,B,C,D).$

Assistant: Multiplying both sides by $(x-2)^3(x-1)$ gives us
$$4x^3 - 20x^2 + 37x  -25 = A(x -2)^3+B(x-1)+C(x -1)(x -2)+D(x-1)(x -2)^2.$$Setting $x=2$ gives $4(8)-20(4)+74-25=B$. Evaluating the expression on the left gives us $B=1$.
Setting $x=1$ gives $4-20+37-25=A(-1)^3$, and so $A=4$.
We still need to find $C$ and $D$. By choosing 2 new values for $x$, we can get two equations which we can solve for $C$ and $D$. We can pick convenient values to make our work easier.
When $x=0$, we get
$$-25=4(-2)^3+(-1)+C(-1)(-2)+D(-1)(-2)^2$$which simplifies to
$$2C-D=8.$$When $x=-1$, we get
$$4(-1)^3-20(-1)^2+37(-1)-25=4(-3)^3+(-2)+C(-2)(-3)+D(-2)(-3)^2$$which simplifies to
$$C-3D=4.$$We can multiply this equation by $2$ and subtract it from the previous one to get $-D+6D=8-2\cdot4=0$ and hence $D=0$. Then $2C=8$ and $C=4$.  Therefore $(A,B,C,D)=(4,1,4,0).$</s>
=======================
The following is a conversation between human and assistant. 

Human: Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be three vectors such that $\|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{c}\| = 2.$  Also, the angle between any two of these vectors is $\arccos \frac{5}{8}.$  Find the volume of the parallelepiped generated by $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}.$

Assistant: Let $\mathbf{p}$ be the projection of $\mathbf{c}$ onto the plane containing $\mathbf{a}$ and $\mathbf{b}.$
[asy]
import three;
size(140);
currentprojection = perspective(6,3,2);
real t = 60, k = Cos(t);
triple A, B, C, O, P, Q;
A = (Cos(t/2),Sin(t/2),0);
B = (Cos(t/2),-Sin(t/2),0);
C = (k/Cos(t/2),0,sqrt(1 - k^2/Cos(t/2)^2));
O = (0,0,0);
P = (k/Cos(t/2),0,0);
Q = k/(k + 1)*A + k/(k + 1)*B;
draw(O--A,Arrow3(6));
draw(O--B,Arrow3(6));
draw(O--C,Arrow3(6));
draw(O--P,Arrow3(6));
draw(C--P,dashed);
label("$\mathbf{a}$", A, S, fontsize(10));
label("$\mathbf{b}$", B, W, fontsize(10));
label("$\mathbf{c}$", C, NW, fontsize(10));
label("$\mathbf{p}$", P, SW, fontsize(10));
[/asy]
Then
\[\mathbf{p} = s \mathbf{a} + t \mathbf{b}\]for some scalars $s$ and $t.$  Let $\mathbf{n}$ be the normal vector to the plane containing $\mathbf{a}$ and $\mathbf{b},$ so
\[\mathbf{c} = \mathbf{p} + u \mathbf{n} = s \mathbf{a} + t \mathbf{b} + u \mathbf{n}\]for some scalar $u.$
Taking the dot product with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{a} + t \mathbf{a} \cdot \mathbf{b} + u \mathbf{a} \cdot \mathbf{n}.\]Note that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 4$ and $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 2 \cdot 2 \cdot \frac{5}{8} = \frac{5}{2}.$  Also, $\mathbf{a} \cdot \mathbf{n} = 0,$ so
\[\frac{5}{2} = 4s + \frac{5t}{2}.\]Similarly, taking the dot product with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{b} + t \mathbf{b} \cdot \mathbf{b} + u \mathbf{b} \cdot \mathbf{n}.\]This reduces to $\frac{5}{2} = \frac{5s}{2} + 4t.$
Solving the equations $\frac{5}{2} = 4s + \frac{5t}{2}$ and $\frac{5}{2} = \frac{5s}{2} + 4t,$ we get $s = t = \frac{5}{13}.$  Hence,
\[\mathbf{p} = \frac{5}{13} (\mathbf{a} + \mathbf{b}).\]Then
\begin{align*}
\|\mathbf{p}\|^2 &= \frac{25}{169} (\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) \\
&= \frac{25}{169} \left( 4 + 2 \cdot \frac{5}{2} + 4 \right) = \frac{25}{13}.
\end{align*}By Pythagoras, the height of the parallelepiped is then given by
\[\sqrt{4 - \|\mathbf{p}\|^2} = \sqrt{4 - \frac{25}{13}} = \sqrt{\frac{27}{13}}.\]The base of the parallelepiped has area $2 \cdot 2 \cdot \sin \left( \arccos \frac{5}{8} \right) = 4 \sqrt{1 - \left( \frac{5}{8} \right)^2} = 4 \sqrt{\frac{39}{64}},$ so the volume of the parallelepiped is
\[\sqrt{\frac{27}{13}} \cdot 4 \sqrt{\frac{39}{64}} = \frac{9}{2}.\]</s>
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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.17s/it]

                                                                                 
Quantizing model.layers blocks :  19%|█▉        | 9/48 [26:07<1:54:05, 175.53s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.69s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.00s/it]

                                                                                 
Quantizing model.layers blocks :  21%|██        | 10/48 [28:59<1:50:22, 174.28s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.08s/it]

                                                                                 
Quantizing model.layers blocks :  23%|██▎       | 11/48 [31:52<1:47:14, 173.90s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.67s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.08s/it]

                                                                                 
Quantizing model.layers blocks :  25%|██▌       | 12/48 [34:44<1:44:02, 173.40s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.10s/it]

                                                                                 
Quantizing model.layers blocks :  27%|██▋       | 13/48 [37:37<1:41:02, 173.22s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.76s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.05s/it]

                                                                                 
Quantizing model.layers blocks :  29%|██▉       | 14/48 [40:29<1:37:56, 172.83s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.71s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.11s/it]

                                                                                 
Quantizing model.layers blocks :  31%|███▏      | 15/48 [43:21<1:34:55, 172.59s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.73s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.73s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.09s/it]

                                                                                 
Quantizing model.layers blocks :  33%|███▎      | 16/48 [46:13<1:31:58, 172.44s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.67s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.79s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.23s/it]

                                                                                 
Quantizing model.layers blocks :  35%|███▌      | 17/48 [49:05<1:29:04, 172.39s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.78s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.73s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.79s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.15s/it]

                                                                                 
Quantizing model.layers blocks :  38%|███▊      | 18/48 [51:58<1:26:11, 172.40s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.80s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.83s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.07s/it]

                                                                                 
Quantizing model.layers blocks :  40%|███▉      | 19/48 [54:50<1:23:17, 172.31s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.83s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.21s/it]

                                                                                 
Quantizing model.layers blocks :  42%|████▏     | 20/48 [57:43<1:20:33, 172.62s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.76s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.73s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.04s/it]

                                                                                 
Quantizing model.layers blocks :  44%|████▍     | 21/48 [1:00:37<1:17:51, 173.01s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.70s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:49<01:06, 16.63s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.78s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.16s/it]

                                                                                 
Quantizing model.layers blocks :  46%|████▌     | 22/48 [1:03:31<1:15:03, 173.22s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.74s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.82s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.84s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.28s/it]

                                                                                 
Quantizing model.layers blocks :  48%|████▊     | 23/48 [1:06:25<1:12:18, 173.54s/it]

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Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.70s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:24<00:33, 16.94s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 17.00s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:42<00:00, 31.36s/it]

                                                                                 
Quantizing model.layers blocks :  50%|█████     | 24/48 [1:09:20<1:09:36, 174.01s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:24, 16.90s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:07, 16.81s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:24<00:33, 16.81s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 16.96s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:42<00:00, 31.40s/it]

                                                                                 
Quantizing model.layers blocks :  52%|█████▏    | 25/48 [1:12:16<1:06:56, 174.64s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:24<00:33, 16.91s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 16.89s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.23s/it]

                                                                                 
Quantizing model.layers blocks :  54%|█████▍    | 26/48 [1:15:11<1:04:06, 174.86s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.71s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 16.87s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.26s/it]

                                                                                 
Quantizing model.layers blocks :  56%|█████▋    | 27/48 [1:18:07<1:01:17, 175.12s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:24<00:33, 16.94s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:17, 17.01s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:42<00:00, 31.43s/it]

                                                                                 
Quantizing model.layers blocks :  58%|█████▊    | 28/48 [1:21:03<58:25, 175.26s/it]  

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:24, 16.89s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 16.90s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.22s/it]

                                                                                 
Quantizing model.layers blocks :  60%|██████    | 29/48 [1:23:57<55:22, 174.85s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.11s/it]

                                                                                 
Quantizing model.layers blocks :  62%|██████▎   | 30/48 [1:26:53<52:36, 175.38s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.81s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.30s/it]

                                                                                 
Quantizing model.layers blocks :  65%|██████▍   | 31/48 [1:29:47<49:33, 174.93s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.78s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.89s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.24s/it]

                                                                                 
Quantizing model.layers blocks :  67%|██████▋   | 32/48 [1:32:41<46:34, 174.66s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:41<00:16, 16.92s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:42<00:00, 31.34s/it]

                                                                                 
Quantizing model.layers blocks :  69%|██████▉   | 33/48 [1:35:35<43:36, 174.45s/it]

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Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.14s/it]

                                                                                 
Quantizing model.layers blocks :  71%|███████   | 34/48 [1:38:28<40:36, 174.05s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.69s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.74s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.22s/it]

                                                                                 
Quantizing model.layers blocks :  73%|███████▎  | 35/48 [1:41:21<37:37, 173.65s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:22, 16.53s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:22<00:33, 16.57s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.66s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.14s/it]

                                                                                 
Quantizing model.layers blocks :  75%|███████▌  | 36/48 [1:44:13<34:38, 173.18s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:22, 16.50s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.68s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:39<00:16, 16.71s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.24s/it]

                                                                                 
Quantizing model.layers blocks :  77%|███████▋  | 37/48 [1:47:06<31:43, 173.03s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.70s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.71s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:50, 16.72s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.77s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.76s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.18s/it]

                                                                                 
Quantizing model.layers blocks :  79%|███████▉  | 38/48 [1:50:01<28:57, 173.71s/it]

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Quantizing layers inside the block:  43%|████▎     | 3/7 [00:49<01:06, 16.57s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.70s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.87s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.25s/it]

                                                                                 
Quantizing model.layers blocks :  81%|████████▏ | 39/48 [1:52:54<26:00, 173.35s/it]

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Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.66s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.75s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:40<00:00, 31.12s/it]

                                                                                 
Quantizing model.layers blocks :  83%|████████▎ | 40/48 [1:55:48<23:09, 173.75s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.71s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.74s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:07<00:50, 16.83s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:24<00:33, 16.81s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.86s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.28s/it]

                                                                                 
Quantizing model.layers blocks :  85%|████████▌ | 41/48 [1:58:43<20:18, 174.08s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:22, 16.56s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:49<01:06, 16.60s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:49, 16.61s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.69s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.75s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.23s/it]

                                                                                 
Quantizing model.layers blocks :  88%|████████▊ | 42/48 [2:01:36<17:23, 173.86s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.77s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:07, 16.76s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:50, 16.69s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.76s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.78s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.13s/it]

                                                                                 
Quantizing model.layers blocks :  90%|████████▉ | 43/48 [2:04:31<14:30, 174.07s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.70s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:49<01:06, 16.58s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:49, 16.64s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.78s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.80s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.16s/it]

                                                                                 
Quantizing model.layers blocks :  92%|█████████▏| 44/48 [2:07:24<11:35, 173.78s/it]

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Quantizing layers inside the block:  14%|█▍        | 1/7 [00:16<01:41, 16.99s/it]

Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.71s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.63s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:49, 16.66s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.75s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.81s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.14s/it]

                                                                                 
Quantizing model.layers blocks :  94%|█████████▍| 45/48 [2:10:18<08:41, 173.88s/it]

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Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.67s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.66s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:50, 16.72s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.83s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.83s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.19s/it]

                                                                                 
Quantizing model.layers blocks :  96%|█████████▌| 46/48 [2:13:13<05:48, 174.07s/it]

Quantizing layers inside the block:   0%|          | 0/7 [00:00<?, ?it/s]

Quantizing layers inside the block:  14%|█▍        | 1/7 [00:16<01:41, 16.88s/it]

Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.73s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.65s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:50, 16.71s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.77s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.82s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.28s/it]

                                                                                 
Quantizing model.layers blocks :  98%|█████████▊| 47/48 [2:16:07<02:54, 174.08s/it]

Quantizing layers inside the block:   0%|          | 0/7 [00:00<?, ?it/s]

Quantizing layers inside the block:  14%|█▍        | 1/7 [00:16<01:40, 16.72s/it]

Quantizing layers inside the block:  29%|██▊       | 2/7 [00:33<01:23, 16.66s/it]

Quantizing layers inside the block:  43%|████▎     | 3/7 [00:50<01:06, 16.72s/it]

Quantizing layers inside the block:  57%|█████▋    | 4/7 [01:06<00:50, 16.72s/it]

Quantizing layers inside the block:  71%|███████▏  | 5/7 [01:23<00:33, 16.79s/it]

Quantizing layers inside the block:  86%|████████▌ | 6/7 [01:40<00:16, 16.85s/it]

Quantizing layers inside the block: 100%|██████████| 7/7 [02:41<00:00, 31.35s/it]

                                                                                 
Quantizing model.layers blocks : 100%|██████████| 48/48 [2:19:01<00:00, 174.21s/it]
Quantizing model.layers blocks : 100%|██████████| 48/48 [2:19:01<00:00, 173.79s/it]
Found modules on cpu/disk. Using Exllama backend requires all the modules to be on GPU. Setting `disable_exllama=True`
2023-11-16 04:30:30.024 | INFO     | __main__:main:67 - Quantizing model...done in 3.609700575878378 hours
2023-11-16 04:30:30.032 | INFO     | __main__:main:68 - Saving model
2023-11-16 04:32:02.594 | INFO     | __main__:main:72 - done