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, stdout

class Solve:
  def __init__(self):
    R = stdin.readline
    W = stdout.write
    n = int(input())
    z, o = 0, 1023
    for i in range(n):
      op, x = R().split()
      x = int(x)
      if op == '|':
        z |= x
        o |= x
      elif op == '&':
        z &= x
        o &= x
      elif op == '^':
        z ^= x
        o ^= x
    ans = {'^' : 0, '|' : 0, '&' : 1023}
    for b in range(10):
      bit = 1 << b
      bitz = z & bit
      bito = o & bit
      if bool(bitz) and not bool(bito):
        ans['^'] |= bit
      elif bool(bitz) and bool(bito):
        ans['|'] |= bit
      elif not bool(bitz) and not bool(bito):
        ans['&'] ^= bit
    W('3\n')
    for op in ans:
      print(op, ans[op])

def main():
  s = Solve()
main()