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.
#include <bits/stdc++.h>
int main() {
  int n, i, j, x, s0 = 0, s1 = 1023, ansxor = 0, ansor = 0, ansand = 1023;
  char s[2];
  scanf("%d", &n);
  while (n--) {
    scanf("%s %d", s, &x);
    switch (s[0]) {
      case '|':
        s0 |= x, s1 |= x;
        break;
      case '^':
        s0 ^= x, s1 ^= x;
        break;
      case '&':
        s0 &= x, s1 &= x;
        break;
      default:
        puts(s);
        puts("Error 1");
        return 0;
    };
  }
  for (i = 0; i < 10; ++i) {
    if ((s0 | (1 << (i))) == s0) {
      if ((s1 | (1 << (i))) != s1)
        ansxor += (1 << (i));
      else
        ansor += (1 << (i)), ansand -= (1 << (i));
    } else {
      if ((s1 | (1 << (i))) != s1)
        ansand -= (1 << (i));
      else
        ;
    }
  }
  printf("3\n& %d\n| %d\n^ %d\n", ansand, ansor, ansxor);
  return 0;
}
