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.
n = int(input())

i0 = 0
i1 = 1023

for i in range(n):
    (cmd, ks) = input().split()
    k = int(ks)
    if cmd == '&':
        i0 &= k
        i1 &= k
    elif cmd == '|':
        i0 |= k
        i1 |= k
    elif cmd == '^':
        i0 ^= k
        i1 ^= k

xor_i = 0
or_i = 0
and_i = 1023

for i in range(10):
    b0 = (i0 >> i) & 1
    b1 = (i1 >> i) & 1
    
    if b0 == 1 and b1 == 0:
        xor_i |= 1 << i
    if b0 == 1 and b1 == 1:
        or_i |= 1 << i
    if b0 == 0 and b1 == 0:
        and_i &= ~(1 << i)

shortened = []
if xor_i != 0:
    shortened.append('^ ' + str(xor_i))
if or_i != 0:
    shortened.append('| ' + str(or_i))
if and_i != 1023:
    shortened.append('& ' + str(and_i))

print(len(shortened))
print("\n".join(shortened))