1463_C. Busy Robot
You have a robot that can move along a number line. At time moment 0 it stands at point 0.

You give n commands to the robot: at time t_i seconds you command the robot to go to point x_i. Whenever the robot receives a command, it starts moving towards the point x_i with the speed of 1 unit per second, and he stops when he reaches that point. However, while the robot is moving, it ignores all the other commands that you give him.

For example, suppose you give three commands to the robot: at time 1 move to point 5, at time 3 move to point 0 and at time 6 move to point 4. Then the robot stands at 0 until time 1, then starts moving towards 5, ignores the second command, reaches 5 at time 6 and immediately starts moving to 4 to execute the third command. At time 7 it reaches 4 and stops there.

You call the command i successful, if there is a time moment in the range [t_i, t_{i + 1}] (i. e. after you give this command and before you give another one, both bounds inclusive; we consider t_{n + 1} = +∞) when the robot is at point x_i. Count the number of successful commands. Note that it is possible that an ignored command is successful.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The next lines describe the test cases.

The first line of a test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of commands.

The next n lines describe the commands. The i-th of these lines contains two integers t_i and x_i (1 ≤ t_i ≤ 10^9, -10^9 ≤ x_i ≤ 10^9) — the time and the point of the i-th command.

The commands are ordered by time, that is, t_i < t_{i + 1} for all possible i.

The sum of n over test cases does not exceed 10^5.

Output

For each testcase output a single integer — the number of successful commands.

Example

Input


8
3
1 5
3 0
6 4
3
1 5
2 4
10 -5
5
2 -5
3 1
4 1
5 1
6 1
4
3 3
5 -3
9 2
12 0
8
1 1
2 -6
7 2
8 3
12 -9
14 2
18 -1
23 9
5
1 -4
4 -7
6 -1
7 -3
8 -7
2
1 2
2 -2
6
3 10
5 5
8 0
12 -4
14 -7
19 -5


Output


1
2
0
2
1
1
0
2

Note

The movements of the robot in the first test case are described in the problem statement. Only the last command is successful.

In the second test case the second command is successful: the robot passes through target point 4 at time 5. Also, the last command is eventually successful.

In the third test case no command is successful, and the robot stops at -5 at time moment 7.

Here are the 0-indexed sequences of the positions of the robot in each second for each testcase of the example. After the cut all the positions are equal to the last one: 

  1. [0, 0, 1, 2, 3, 4, 5, 4, 4, ...] 
  2. [0, 0, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -5, ...] 
  3. [0, 0, 0, -1, -2, -3, -4, -5, -5, ...] 
  4. [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 2, 2, 2, 1, 0, 0, ...] 
  5. [0, 0, 1, 0, -1, -2, -3, -4, -5, -6, -6, -6, -6, -7, -8, -9, -9, -9, -9, -8, -7, -6, -5, -4, -3, -2, -1, -1, ...] 
  6. [0, 0, -1, -2, -3, -4, -4, -3, -2, -1, -1, ...] 
  7. [0, 0, 1, 2, 2, ...] 
  8. [0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -7, ...] 
```python
import sys
import bisect
input=sys.stdin.readline
t=int(input())
for you in range(t):
    n=int(input())
    l=[]
    for i in range(n):
        z=input().split()
        u=int(z[0])
        v=int(z[1])
        l.append((u,v))
    l.append((10**18,0))
    counter=0
    currtime=0
    currpos=0
    reqtime=0
    reqpos=0
    for i in range(n):
        if(reqtime<=l[i][0]):
            #print("yes")
            currtime=l[i][0]
            currpos=reqpos
            reqtime=currtime+abs(currpos-l[i][1])
            reqpos=l[i][1]
            if(abs(reqpos-currpos)+currtime<=l[i+1][0]):
                counter+=1
                #print("done")
        else:
            #print("no")
            if((reqpos-currpos)*(l[i][1]-currpos)<0):
                continue
            if(abs(reqpos-currpos)<abs(l[i][1]-currpos)):
                continue
            z=currtime+abs(l[i][1]-currpos)
            #print(z)
            if(z>=l[i][0] and z<=l[i+1][0]):
                counter+=1
                #print("done")
    print(counter)
```