#include <bits/stdc++.h>
using namespace std;
int n, b[20];
bool wie[20];
vector<int> x[20], a[20], o[20];
int co1[20], co2[20], co3[20];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    char c;
    int y;
    cin >> c >> y;
    if (c == '^') {
      for (int j = 0; j <= 11; j++) {
        if (y & (1 << j)) {
          x[j].push_back(1);
          if (wie[j]) b[j] ^= x[j].back();
        }
      }
    }
    if (c == '|') {
      for (int j = 0; j <= 11; j++) {
        if (y & (1 << j)) {
          wie[j] = 1;
          b[j] = 1;
          o[j].push_back(1);
        }
      }
    }
    if (c == '&') {
      for (int j = 0; j <= 11; j++) {
        if (!(y & (1 << j))) {
          wie[j] = 1;
          b[j] = 0;
          a[j].push_back(0);
        }
      }
    }
  }
  for (int i = 0; i < 10; i++) {
    if (wie[i]) {
      if (b[i] == 0) {
        co1[i] = 0;
        co2[i] = 0;
        co3[i] = 0;
      } else {
        co1[i] = 0;
        co2[i] = 1;
        co3[i] = 1;
      }
    } else {
      co1[i] = ((int)x[i].size()) % 2;
      co2[i] = 1;
      co3[i] = 0;
    }
  }
  cout << 3 << "\n";
  int p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co1[i];
    p *= 2;
  }
  cout << "^ " << l << "\n";
  p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co2[i];
    p *= 2;
  }
  cout << "& " << l << "\n";
  p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co3[i];
    p *= 2;
  }
  cout << "| " << l << "\n";
  return 0;
}

### Prompt

Please formulate 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, b[20];
bool wie[20];
vector<int> x[20], a[20], o[20];
int co1[20], co2[20], co3[20];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    char c;
    int y;
    cin >> c >> y;
    if (c == '^') {
      for (int j = 0; j <= 11; j++) {
        if (y & (1 << j)) {
          x[j].push_back(1);
          if (wie[j]) b[j] ^= x[j].back();
        }
      }
    }
    if (c == '|') {
      for (int j = 0; j <= 11; j++) {
        if (y & (1 << j)) {
          wie[j] = 1;
          b[j] = 1;
          o[j].push_back(1);
        }
      }
    }
    if (c == '&') {
      for (int j = 0; j <= 11; j++) {
        if (!(y & (1 << j))) {
          wie[j] = 1;
          b[j] = 0;
          a[j].push_back(0);
        }
      }
    }
  }
  for (int i = 0; i < 10; i++) {
    if (wie[i]) {
      if (b[i] == 0) {
        co1[i] = 0;
        co2[i] = 0;
        co3[i] = 0;
      } else {
        co1[i] = 0;
        co2[i] = 1;
        co3[i] = 1;
      }
    } else {
      co1[i] = ((int)x[i].size()) % 2;
      co2[i] = 1;
      co3[i] = 0;
    }
  }
  cout << 3 << "\n";
  int p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co1[i];
    p *= 2;
  }
  cout << "^ " << l << "\n";
  p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co2[i];
    p *= 2;
  }
  cout << "& " << l << "\n";
  p = 1, l = 0;
  for (int i = 0; i < 10; i++) {
    l += p * co3[i];
    p *= 2;
  }
  cout << "| " << l << "\n";
  return 0;
}

```