Solve in Python:
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 ≤ 5·10^5) — 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 x_{i} 0 ≤ x_{i} ≤ 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
& 1
& 3
& 5

Output
1
& 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.
n = int(input())
queries = list(input().split() for _ in range(n))

a, b = 0, (1<<10) - 1
for c, x in queries:
    x = int(x)
    if c == '|':
        a, b = a | x, b | x
    elif c == '&':
        a, b = a & x, b & x
    elif c == '^':
        a, b = a ^ x, b ^ x

x, y, z = 0, (1<<10) - 1, 0
for i in range(10):
    a_i = (a >> i) & 1
    b_i = (b >> i) & 1
    if a_i and b_i:
        x |= (1 << i)
    if not a_i and not b_i:
        y ^= (1 << i)
    if a_i and not b_i:
        z ^= (1 << i)

print(3)
print('|', x)
print('&', y)
print('^', z)