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.
import sys, math


def main():
    ops = getops(sys.stdin)

    n1 = 0
    n2 = 1023
    for op in ops:
        n1 = calc(op, n1)
        n2 = calc(op, n2)

    if n2 == 1023 and n1 == 0:
        print(0)
    else:
        and_ = 1023
        or_ = 0
        xor_ = 0
        for i in range(10):
            b1 = n1 % 2
            b2 = n2 % 2
            n1 = n1 // 2
            n2 = n2 // 2
            if b1 == 1 and b2 == 1:
                or_ = or_ | 2**i
            elif b1 == 1 and b2 == 0:
                xor_ = xor_ | 2**i
            elif b1 == 0 and b2 == 0:
                and_ = and_ - 2**i

        print(3)
        print('&', and_)
        print('|', or_)
        print('^', xor_)


def getops(fh):
    n = int(fh.readline())
    ops = []
    for i in range(n):
        opi, numi = fh.readline().split()
        numi = int(numi)
        ops.append((opi, numi))

    return ops


def calc(op, res):
    opi, numi = op
    if opi == '|':
        res = res | numi
    elif opi == '&':
        res = res & numi
    else:
        res = res ^ numi
    return res


if __name__ == "__main__":
    main()
