#include <bits/stdc++.h>
using namespace std;
bool yes(int x, int i) {
  int y = x;
  y = y >> i;
  y = y & 1;
  return y > 0;
}
int main() {
  int n;
  cin >> n;
  int x1 = 0, x2 = 1023;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    int k;
    cin >> k;
    switch (c) {
      case '&':
        x1 = x1 & k;
        x2 = x2 & k;
        break;
      case '|':
        x1 = x1 | k;
        x2 = x2 | k;
        break;
      case '^':
        x1 = x1 ^ k;
        x2 = x2 ^ k;
        break;
    }
  }
  int xa = 1023, xo = 0, xx = 0;
  for (int i = 0; i < 10; i++) {
    if (yes(x1, i) && yes(x2, i)) {
      xo += (1 << i);
    }
    if (!yes(x1, i) && !yes(x2, i)) {
      xa -= (1 << i);
    }
    if (yes(x1, i) && !yes(x2, i)) {
      xx += (1 << i);
    }
  }
  int k = 0;
  if (xa != 1023) k++;
  if (xo != 0) k++;
  if (xx != 0) k++;
  cout << k << endl;
  if (xa != 1023) cout << "& " << xa << endl;
  if (xo != 0) cout << "| " << xo << endl;
  if (xx != 0) cout << "^ " << xx << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
bool yes(int x, int i) {
  int y = x;
  y = y >> i;
  y = y & 1;
  return y > 0;
}
int main() {
  int n;
  cin >> n;
  int x1 = 0, x2 = 1023;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    int k;
    cin >> k;
    switch (c) {
      case '&':
        x1 = x1 & k;
        x2 = x2 & k;
        break;
      case '|':
        x1 = x1 | k;
        x2 = x2 | k;
        break;
      case '^':
        x1 = x1 ^ k;
        x2 = x2 ^ k;
        break;
    }
  }
  int xa = 1023, xo = 0, xx = 0;
  for (int i = 0; i < 10; i++) {
    if (yes(x1, i) && yes(x2, i)) {
      xo += (1 << i);
    }
    if (!yes(x1, i) && !yes(x2, i)) {
      xa -= (1 << i);
    }
    if (yes(x1, i) && !yes(x2, i)) {
      xx += (1 << i);
    }
  }
  int k = 0;
  if (xa != 1023) k++;
  if (xo != 0) k++;
  if (xx != 0) k++;
  cout << k << endl;
  if (xa != 1023) cout << "& " << xa << endl;
  if (xo != 0) cout << "| " << xo << endl;
  if (xx != 0) cout << "^ " << xx << endl;
  return 0;
}

```