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 CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  int a = 0, b = 1023, n, x;
  char c;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("\n%c %d", &c, &x);
    switch (c) {
      case '|':
        a |= x;
        b |= x;
        break;
      case '&':
        a &= x;
        b &= x;
        break;
      case '^':
        a ^= x;
        b ^= x;
        break;
    }
  }
  int a1 = 0, a2 = 0;
  for (int i = 0, k = 1; i < 10; i += 1, k *= 2) {
    if (!(a & k) && !(b & k)) {
    } else if (!(a & k) && (b & k)) {
      a1 += k;
    } else if ((a & k) && !(b & k)) {
      a1 += k;
      a2 += k;
    } else if ((a & k) && (b & k)) {
      a2 += k;
    }
  }
  printf("2\n& %d\n^ %d\n", a1, a2);
  return 0;
}