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 PYTHON):

Solution:
LEN = 10
mapping = [[0, 1] for _ in range(LEN)]
n = int(raw_input())
for _ in range(n):
    cmd = raw_input().split()
    val = int(cmd[1])
    if cmd[0] == '|':
        for i in range(LEN):
            cur_digit = ((val & (1 << i) ) != 0)
            if cur_digit:
                mapping[i][0], mapping[i][1] = 1, 1
    elif cmd[0] == '&':
        for i in range(LEN):
            cur_digit = ((val & (1 << i) ) != 0)
            if not cur_digit:
                mapping[i][0], mapping[i][1] = 0, 0
    else:
        for i in range(LEN):
            cur_digit = ((val & (1 << i) ) != 0)
            if cur_digit:
                mapping[i][0], mapping[i][1] = mapping[i][0] ^ 1, mapping[i][1] ^ 1

# construct answer
# &
and_val = (1 << LEN) - 1
for i in range(LEN):
    if mapping[i][0] == 0 and mapping[i][1] == 0:
        and_val = and_val ^ (1 << i)

or_val = 0
for i in range(LEN):
    if mapping[i][0] == 1 and mapping[i][1] == 1:
        or_val = or_val | (1 << i)

xor_val = 0
for i in range(LEN):
    if mapping[i][0] == 1 and mapping[i][1] == 0:
        xor_val = xor_val | (1 << i)

print 3
print '& %d' % and_val
print '| %d' % or_val
print '^ %d' % xor_val