878_A. Short Program
Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.

In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.

Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023.

Input

The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines.

Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.

Output

Output an integer k (0 ≤ k ≤ 5) — the length of your program.

Next k lines must contain commands in the same format as in the input.

Examples

Input

3
| 3
^ 2
| 1


Output

2
| 3
^ 2


Input

3
&amp; 1
&amp; 3
&amp; 5


Output

1
&amp; 1


Input

3
^ 1
^ 2
^ 3


Output

0

Note

You can read about bitwise operations in <https://en.wikipedia.org/wiki/Bitwise_operation>.

Second sample:

Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.
```python
n = int(input())

zero = 0
ones = 1023

for _ in range(n):
    op, num = input().split()
    
    num = int(num)
    
    if op == '&':
        zero &= num
        ones &= num
    elif op == '|':
        zero |= num
        ones |= num
    else:
        zero ^= num
        ones ^= num

and_bits = 0
or_bits  = 0
xor_bits = 0

for i in range(10):
    z = (zero >> i) & 1
    o = (ones >> i) & 1
    
    if z == 0 and o == 0:
        and_bits |= 0 << i
        or_bits  |= 0 << i
        xor_bits |= 0 << i
    elif z == 0 and o == 1:
        and_bits |= 1 << i
        or_bits  |= 0 << i
        xor_bits |= 0 << i
    elif z == 1 and o == 0:
        and_bits |= 1 << i
        or_bits  |= 0 << i
        xor_bits |= 1 << i
    else:
        and_bits |= 1 << i
        or_bits  |= 1 << i
        xor_bits |= 0 << i

print(3)
print('& {}'.format(and_bits))
print('| {}'.format(or_bits))
print('^ {}'.format(xor_bits))
```