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>
char a[500010];
int b[500010];
int res[10][2];
int op[10][4];
int main() {
  int n;
  scanf("%d", &n);
  char tmp[10];
  for (int i = 1; i <= n; i++) {
    scanf("%s", tmp);
    a[i] = tmp[0];
    scanf("%d", &b[i]);
  }
  int x = 1;
  for (int i = 0; i < 10; i++) {
    int y = x;
    int z = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j] == '|') {
        z |= b[j];
        y |= b[j];
      } else if (a[j] == '&') {
        z &= b[j];
        y &= b[j];
      } else {
        z ^= b[j];
        y ^= b[j];
      }
    }
    y >>= i;
    z >>= i;
    res[i][1] = y % 2;
    res[i][0] = z % 2;
    x *= 2;
  }
  int aa = 0, bb = 0, cc = 0;
  for (int i = 0; i < 10; i++) {
    if (res[i][1] == 1 && res[i][0] == 0) {
      op[i][0] = 1;
    } else if (res[i][1] == 1 && res[i][0] == 1) {
      op[i][0] = 1;
      op[i][1] = 1;
    } else if (res[i][1] == 0 && res[i][0] == 0) {
      op[i][0] = 0;
    } else {
      op[i][0] = 1;
      op[i][2] = 1;
    }
  }
  for (int i = 9; i >= 0; i--) {
    aa *= 2;
    bb *= 2;
    cc *= 2;
    aa += op[i][0];
    bb += op[i][1];
    cc += op[i][2];
  }
  printf("%d\n", 3);
  printf("& %d\n", aa);
  printf("| %d\n", bb);
  printf("^ %d\n", cc);
  return 0;
}
