#include <bits/stdc++.h>
using namespace std;
int a[2][10];
int main() {
  for (int i = 0; i < 10; i++) a[1][i] = 1;
  int n, m;
  char c;
  scanf("%d", &n);
  getchar();
  for (int i = 0; i < n; i++) {
    scanf("%c %d", &c, &m);
    getchar();
    if (c == '&') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] &= ((m >> j) & 1);
    }
    if (c == '|') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] |= ((m >> j) & 1);
    }
    if (c == '^') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] ^= ((m >> j) & 1);
    }
  }
  int x = 0, y = 0, z = 0;
  for (int i = 9; i >= 0; i--) {
    if (a[0][i] == 0 && a[1][i] == 1) {
      x = x * 2 + 1;
      y = y * 2 + 0;
      z = z * 2 + 0;
    } else if (a[0][i] == 1 && a[1][i] == 0) {
      x = x * 2 + 1;
      y = y * 2 + 0;
      z = z * 2 + 1;
    } else if (a[0][i] == 0 && a[1][i] == 0) {
      x = x * 2 + 0;
      y = y * 2 + 0;
      z = z * 2 + 0;
    } else {
      x = x * 2 + 1;
      y = y * 2 + 1;
      z = z * 2 + 0;
    }
  }
  printf("3\n");
  printf("& %d\n", x);
  printf("| %d\n", y);
  printf("^ %d\n", z);
  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 a[2][10];
int main() {
  for (int i = 0; i < 10; i++) a[1][i] = 1;
  int n, m;
  char c;
  scanf("%d", &n);
  getchar();
  for (int i = 0; i < n; i++) {
    scanf("%c %d", &c, &m);
    getchar();
    if (c == '&') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] &= ((m >> j) & 1);
    }
    if (c == '|') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] |= ((m >> j) & 1);
    }
    if (c == '^') {
      for (int j = 0; j < 10; j++)
        for (int k = 0; k < 2; k++) a[k][j] ^= ((m >> j) & 1);
    }
  }
  int x = 0, y = 0, z = 0;
  for (int i = 9; i >= 0; i--) {
    if (a[0][i] == 0 && a[1][i] == 1) {
      x = x * 2 + 1;
      y = y * 2 + 0;
      z = z * 2 + 0;
    } else if (a[0][i] == 1 && a[1][i] == 0) {
      x = x * 2 + 1;
      y = y * 2 + 0;
      z = z * 2 + 1;
    } else if (a[0][i] == 0 && a[1][i] == 0) {
      x = x * 2 + 0;
      y = y * 2 + 0;
      z = z * 2 + 0;
    } else {
      x = x * 2 + 1;
      y = y * 2 + 1;
      z = z * 2 + 0;
    }
  }
  printf("3\n");
  printf("& %d\n", x);
  printf("| %d\n", y);
  printf("^ %d\n", z);
  return 0;
}

```