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.
def cal(num1, op, num2):
    if op == '^':
        return num1 ^ num2
    if op == '&':
        return num1 & num2
    if op == '|':
        return num1 | num2

def solve():
    n = int(raw_input())
    x0 = 0
    x1 = (1 << 10) - 1

    for i in xrange(n):
        inp = raw_input().split()
        op = inp[0]
        other = int(inp[1])
        x0 = cal(x0, op, other)
        x1 = cal(x1, op, other)

    ops = {
        '|': 0,
        '^': 0,
        '&': (1 << 10) - 1,
    }
    for pos in xrange(10):
        x0_d = x0 & (1 << pos)
        x1_d = x1 & (1 << pos)
        if x0_d == 0 and x1_d == 0:
            ops['&'] -= (1 << pos)
        elif x0_d > 0 and x1_d > 0:
            ops['|'] += (1 << pos)
        elif x0_d > 0 and x1_d == 0:
            ops['^'] += (1 << pos)

    print('3')
    for op, other in ops.items():
        print('{} {}'.format(op, other))


if __name__ == '__main__':
    solve()
