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.
q = int(input())

def calc(f):
    ret1 = 0
    ret2 = 1023
    for op in f:
        if op[0] == '&':
            ret1 &= op[1]
            ret2 &= op[1]
        elif op[0] == '|':
            ret1 |= op[1]
            ret2 |= op[1]
        elif op[0] == '^':
            ret1 ^= op[1]
            ret2 ^= op[1]

    return (ret1, ret2)

f = list()
for i in range(q):
    oper, n = input().split()
    n = int(n)
    f.append((oper, n))

op0, op1 = calc(f)

op0 = ('0000000000' + bin(op0)[2:])[-10:][::-1]
op1 = ('0000000000' + bin(op1)[2:])[-10:][::-1]

o = 0
x = 0

for i in range(10):
    if (op0[i] == '1') and (op1[i] == '1'):
        o += 2**i

    elif (op0[i] == '1') and (op1[i] == '0'):
        x += 2**i

    elif (op0[i] == '0') and (op1[i] == '0'):
        o += 2**i
        x += 2**i

print(2)
print('| ' + str(o))
print('^ ' + str(x))
