#include <bits/stdc++.h>
const int N = 500 * 1000, of_bits = 10;
int n, where_set[of_bits], consts[N], one, two, three;
char str[20], op[N], one_xor[of_bits], val[of_bits];
int main() {
  std::scanf("%d", &n);
  for (int i = 0; i < of_bits; i++) val[i] = -1;
  for (int i = 0; i < n; i++) {
    std::scanf("%s%d", str, &consts[i]);
    op[i] = str[0];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (op[i] == '&') {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && !((1 << j) & consts[i])) val[j] = one_xor[j];
    } else if (op[i] == '|') {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && ((1 << j) & consts[i]))
          val[j] = (one_xor[j] + 1) % 2;
    } else {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && ((1 << j) & consts[i]))
          one_xor[j] = (one_xor[j] + 1) % 2;
    }
  }
  for (int j = 0; j < of_bits; j++)
    if (val[j] == 1) one |= (1 << j);
  for (int j = 0; j < of_bits; j++)
    if (val[j] != 0) two |= (1 << j);
  for (int j = 0; j < of_bits; j++)
    if (val[j] == -1 && one_xor[j]) three |= (1 << j);
  std::printf("3\n");
  std::printf("| %d\n", one);
  std::printf("& %d\n", two);
  std::printf("^ %d\n", three);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int N = 500 * 1000, of_bits = 10;
int n, where_set[of_bits], consts[N], one, two, three;
char str[20], op[N], one_xor[of_bits], val[of_bits];
int main() {
  std::scanf("%d", &n);
  for (int i = 0; i < of_bits; i++) val[i] = -1;
  for (int i = 0; i < n; i++) {
    std::scanf("%s%d", str, &consts[i]);
    op[i] = str[0];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (op[i] == '&') {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && !((1 << j) & consts[i])) val[j] = one_xor[j];
    } else if (op[i] == '|') {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && ((1 << j) & consts[i]))
          val[j] = (one_xor[j] + 1) % 2;
    } else {
      for (int j = 0; j < of_bits; j++)
        if (val[j] == -1 && ((1 << j) & consts[i]))
          one_xor[j] = (one_xor[j] + 1) % 2;
    }
  }
  for (int j = 0; j < of_bits; j++)
    if (val[j] == 1) one |= (1 << j);
  for (int j = 0; j < of_bits; j++)
    if (val[j] != 0) two |= (1 << j);
  for (int j = 0; j < of_bits; j++)
    if (val[j] == -1 && one_xor[j]) three |= (1 << j);
  std::printf("3\n");
  std::printf("| %d\n", one);
  std::printf("& %d\n", two);
  std::printf("^ %d\n", three);
  return 0;
}

```