Problem: 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.

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout

n = int(stdin.readline())
a = list()
y = 0

for i in range(n):
    oper, x = stdin.readline().split()
    x = int(x)
    a.append((oper, x))
    if oper == '&':
        y &= x
    elif oper == '|':
        y |= x
    elif oper == '^':
        y ^= x

AND = (1<<10) - 1
OR = 0
XOR = 0

for k in range(10):
    cur = 1 << k
    for comm in a:
        oper, x = comm
        if oper == '&':
            cur &= x
        elif oper == '|':
            cur |= x
        elif oper == '^':
            cur ^= x
    res0 = y & (1<<k)
    res1 = cur & (1<<k)
    if not res0 and not res1:
        AND ^= (1<<k)
    elif not res0 and res1:
        pass
    elif res0 and not res1:
        XOR ^= (1<<k)
    else:
        OR ^= (1<<k)

print(3)
print('& ', AND)
print('| ', OR)
print('^ ', XOR)