#include <bits/stdc++.h>
using namespace std;
int n, bt[500010][12], ans[15][5], res[10];
char op[500010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) {
    char s[2];
    scanf("%s", s);
    op[i] = s[0];
    int num;
    scanf("%d", &num);
    for (int j = 0; j < 9 + 1; j++) {
      if (num & 1)
        bt[i][j] = 1;
      else
        bt[i][j] = 0;
      num /= 2;
    }
  }
  for (int i = 1; i < n + 1; i++) {
    for (int j = 0; j < 9 + 1; j++) {
      if (op[i] == '&') {
        if (bt[i][j] == 0) {
          ans[j][0] = 1;
          for (int k = 1; k < 3 + 1; k++) ans[j][k] = 0;
        }
      } else if (op[i] == '|') {
        if (bt[i][j] == 1) {
          ans[j][1] = 1;
          ans[j][0] = ans[j][2] = ans[j][3] = 0;
        }
      } else {
        if (bt[i][j] == 1) {
          if (ans[j][0] || ans[j][1])
            swap(ans[j][0], ans[j][1]);
          else
            ans[j][3] ^= 1;
        }
      }
    }
  }
  for (int i = 0; i < 9 + 1; i++) {
    if (ans[i][0])
      ;
    else
      res[0] |= (1 << i);
    if (ans[i][1]) res[1] |= (1 << i);
    if (ans[i][3]) res[3] |= (1 << i);
  }
  printf("3\n");
  printf("& %d\n", res[0]);
  printf("| %d\n", res[1]);
  printf("^ %d", res[3]);
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, bt[500010][12], ans[15][5], res[10];
char op[500010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) {
    char s[2];
    scanf("%s", s);
    op[i] = s[0];
    int num;
    scanf("%d", &num);
    for (int j = 0; j < 9 + 1; j++) {
      if (num & 1)
        bt[i][j] = 1;
      else
        bt[i][j] = 0;
      num /= 2;
    }
  }
  for (int i = 1; i < n + 1; i++) {
    for (int j = 0; j < 9 + 1; j++) {
      if (op[i] == '&') {
        if (bt[i][j] == 0) {
          ans[j][0] = 1;
          for (int k = 1; k < 3 + 1; k++) ans[j][k] = 0;
        }
      } else if (op[i] == '|') {
        if (bt[i][j] == 1) {
          ans[j][1] = 1;
          ans[j][0] = ans[j][2] = ans[j][3] = 0;
        }
      } else {
        if (bt[i][j] == 1) {
          if (ans[j][0] || ans[j][1])
            swap(ans[j][0], ans[j][1]);
          else
            ans[j][3] ^= 1;
        }
      }
    }
  }
  for (int i = 0; i < 9 + 1; i++) {
    if (ans[i][0])
      ;
    else
      res[0] |= (1 << i);
    if (ans[i][1]) res[1] |= (1 << i);
    if (ans[i][3]) res[3] |= (1 << i);
  }
  printf("3\n");
  printf("& %d\n", res[0]);
  printf("| %d\n", res[1]);
  printf("^ %d", res[3]);
  return 0;
}

```