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

data = stdin.readlines()

remain = -10
invert = -20

m = [remain] * 10

for i, st in enumerate(data):
    if i == 0: continue
    op, x = st.strip().split()
    x = int(x)

    for j in range(10):
        bit = x % 2
        if op == '&' and bit == 0:
            m[j] = 0
        elif op == '|' and bit == 1:
            m[j] = 1
        elif op == '^' and bit == 1:
            if m[j] == 0:
                m[j] = 1
            elif m[j] == 1:
                m[j] = 0
            elif m[j] == remain:
                m[j] = invert
            elif m[j] == invert:
                m[j] = remain
        x = x // 2

print(3)
and_const = 0
for i, v in enumerate(m):
    if v != 0:
        and_const += 2**i

or_const = 0
for i, v in enumerate(m):
    if v == 1:
        or_const += 2**i

xor_const = 0
for i, v in enumerate(m):
    if v == invert:
        xor_const += 2**i

print('&', and_const)
print('|', or_const)
print('^', xor_const)