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.
from sys import stdin

rstrs = lambda: [str(x) for x in input().split()]
rstrs_2d = lambda n: [rstrs() for _ in range(n)]
get_bit = lambda x, i: (x >> i) & 1


def fast2():
    import os, sys, atexit
    from cStringIO import StringIO as BytesIO
    # range = xrange
    sys.stdout = BytesIO()
    atexit.register(lambda: os.write(1, sys.stdout.getvalue()))
    return BytesIO(os.read(0, os.fstat(0).st_size)).readline


input = fast2()
n = int(input())
a, out = rstrs_2d(n), [['|', ['0'] * 10], ['^', ['0'] * 10], ['&', ['1'] * 10]]

for i in range(10):
    all = []
    for j in range(2):
        cur = j
        for op, num in a:
            if op == '|':
                cur |= get_bit(int(num), i)
            elif op == '&':
                cur &= get_bit(int(num), i)
            else:
                cur ^= get_bit(int(num), i)
        all.append(cur)

    if all[0] == all[1]:
        if all[0]:
            out[0][1][-i - 1] = '1'
        else:
            out[2][1][-i - 1] = '0'
    elif all[0] == 1 and all[1] == 0:
        out[1][1][-i - 1] = '1'

print(3)
for i, j in out:
    print('%s %d' % (i, int(''.join(j), 2)))
