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>
using namespace std;
int one[15], zero[15], digits[15];
int main() {
  int number, digit, andn, orn, xorn, n;
  char command;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < 10; i++) {
      zero[i] = 0;
      one[i] = 1;
    }
    for (int i = 0; i < n; i++) {
      getchar();
      scanf("%c %d", &command, &number);
      digit = 0;
      for (int i = 0; i < 10; i++) {
        digits[digit++] = number % 2;
        number /= 2;
      }
      switch (command) {
        case '&':
          for (int i = 0; i < 10; i++) {
            zero[i] &= digits[i];
            one[i] &= digits[i];
          }
          break;
        case '|':
          for (int i = 0; i < 10; i++) {
            zero[i] |= digits[i];
            one[i] |= digits[i];
          }
          break;
        case '^':
          for (int i = 0; i < 10; i++) {
            zero[i] ^= digits[i];
            one[i] ^= digits[i];
          }
          break;
      }
    }
    andn = orn = xorn = 0;
    for (int i = 0; i < 10; i++) {
      if (one[i] == 1 && zero[i] == 0) {
        andn += (1 << i);
      } else if (zero[i] == one[i] && zero[i] == 1) {
        orn += (1 << i);
        andn += (1 << i);
      } else if (zero[i] == one[i] && zero[i] == 0) {
        orn += (1 << i);
      } else {
        andn += (1 << i);
        xorn += (1 << i);
      }
    }
    printf("3\n| %d\n& %d\n^ %d\n", orn, andn, xorn);
  }
  return 0;
}