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 N;
int save[10];
int main() {
  for (int n = 0; n < 10; n++) save[n] = 3;
  scanf("%d", &N);
  for (int n = 0; n < N; n++) {
    char c;
    int d;
    scanf(" %c%d", &c, &d);
    if (c == '|') {
      for (int t = 1, k = 0; t < 1024; t *= 2, k++) {
        if (d & t) save[k] = 1;
      }
    } else if (c == '&') {
      for (int t = 1, k = 0; t < 1024; t *= 2, k++) {
        if (!(d & t)) save[k] = 0;
      }
    } else {
      for (int t = 1, k = 0; t < 1024; t *= 2, k++) {
        if (d & t) {
          if (save[k] > 1)
            save[k] = 5 - save[k];
          else
            save[k] = 1 - save[k];
        }
      }
    }
  }
  printf("3\n");
  int tmp = 0;
  for (int k = 0; k < 10; k++)
    if (save[k] == 1) tmp |= (1 << k);
  printf("| %d\n", tmp);
  tmp = 1023;
  for (int k = 0; k < 10; k++)
    if (save[k] == 0) tmp ^= (1 << k);
  printf("& %d\n", tmp);
  tmp = 0;
  for (int k = 0; k < 10; k++)
    if (save[k] == 2) tmp |= (1 << k);
  printf("^ %d\n", tmp);
  return 0;
}