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 operator import __or__, __and__, __xor__
from sys import stdin, stdout
n, b, c = int(stdin.readline()), 0, 1023
m = {'|': __or__, '&': __and__, '^': __xor__}
for i in range(n):
    t, v = [i for i in stdin.readline().split()]
    b = m[t](b, int(v))
    c = m[t](c, int(v))
x, o, a = 0, 0, 1023
for i in range(10):
    if ((b >> i) & 1) and ((c >> i) & 1):
        o |= 1 << i
    elif not ((b >> i) & 1) and not ((c >> i) & 1):
        a -= 1 << i
    elif ((b >> i) & 1) and not ((c >> i) & 1):
        x |= 1 << i
stdout.write('3\n| ' + str(o) + '\n^ ' + str(x) + '\n& ' + str(a))