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.
#   ==========     //\\       //||     ||====//||
#       ||        //  \\        ||     ||   // ||
#       ||       //====\\       ||     ||  //  ||
#       ||      //      \\      ||     || //   ||
#   ========== //        \\  ========  ||//====|| 
#  code

def solve():
    n = int(input())
    a = 0
    b = 1023

    for i in range(n):
        op, k = input().split()
        k = int(k)
        if op == '|':
            a |= k
            b |= k
        if op == '^':
            a ^= k
            b ^= k
        if op == '&':
            a &= k
            b &= k
    
    p = 1
    print(3)
    aa = a
    bb = b
    ans = 0
    for i in range(10):
        if aa % 2 == 1 and bb % 2 == 0:
            ans += p
        p *= 2
        aa //= 2
        bb //= 2
    print('^', ans)

    p = 1
    ans = 1023
    aa = a
    bb = b
    for i in range(10):
        if aa % 2 == 0 and bb % 2 == 0:
            ans -= p
        p *= 2
        aa //= 2
        bb //= 2
    print('&', ans)

    p = 1
    ans = 0
    aa = a
    bb = b
    for i in range(10):
        if aa % 2 == 1 and bb % 2 == 1:
            ans += p
        p *= 2
        aa //= 2
        bb //= 2
    print('|', ans)
    return

def main():
    t = 1
    # t = int(input())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()