#include <bits/stdc++.h>
using namespace std;
const int MAX = 5e5 + 5;
int XOR = 0, AND = 1, OR = 2;
int N;
vector<pair<int, int> > v;
int A[MAX], B[3];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    char x;
    cin >> x;
    int y;
    cin >> y;
    if (x == '^')
      v.push_back({XOR, y});
    else if (x == '&')
      v.push_back({AND, y});
    else if (x == '|')
      v.push_back({OR, y});
  }
  for (int i = 0; i < 10; ++i) {
    int swap = 1, same = 1, one = 0, zero = 0;
    for (int j = 0; j <= 1; ++j) {
      int prev = j;
      for (int k = 0; k < v.size(); ++k) {
        if (v[k].first == XOR) {
          if (v[k].second & (1LL << i))
            prev ^= 1;
          else
            ;
        }
        if (v[k].first == AND) {
          if (v[k].second & (1LL << i))
            ;
          else
            prev = 0;
        }
        if (v[k].first == OR) {
          if (v[k].second & (1LL << i))
            prev = 1;
          else
            ;
        }
      }
      if (prev == j)
        swap = 0;
      else
        same = 0;
      if (prev == 0)
        ++zero;
      else
        ++one;
    }
    if (same)
      A[i] = 0;
    else if (swap)
      A[i] = 1;
    else if (zero == 2)
      A[i] = 2;
    else
      A[i] = 3;
  }
  for (int i = 0; i < 10; ++i) {
    if (A[i] == 0)
      B[1] |= (1LL << i);
    else if (A[i] == 1)
      B[0] |= (1LL << i), B[1] |= (1LL << i);
    else if (A[i] == 2)
      ;
    else if (A[i] == 3)
      B[1] |= (1LL << i), B[2] |= (1LL << i);
  }
  cout << 3 << endl;
  cout << "^ " << B[0] << endl;
  cout << "& " << B[1] << endl;
  cout << "| " << B[2] << endl;
  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;
const int MAX = 5e5 + 5;
int XOR = 0, AND = 1, OR = 2;
int N;
vector<pair<int, int> > v;
int A[MAX], B[3];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    char x;
    cin >> x;
    int y;
    cin >> y;
    if (x == '^')
      v.push_back({XOR, y});
    else if (x == '&')
      v.push_back({AND, y});
    else if (x == '|')
      v.push_back({OR, y});
  }
  for (int i = 0; i < 10; ++i) {
    int swap = 1, same = 1, one = 0, zero = 0;
    for (int j = 0; j <= 1; ++j) {
      int prev = j;
      for (int k = 0; k < v.size(); ++k) {
        if (v[k].first == XOR) {
          if (v[k].second & (1LL << i))
            prev ^= 1;
          else
            ;
        }
        if (v[k].first == AND) {
          if (v[k].second & (1LL << i))
            ;
          else
            prev = 0;
        }
        if (v[k].first == OR) {
          if (v[k].second & (1LL << i))
            prev = 1;
          else
            ;
        }
      }
      if (prev == j)
        swap = 0;
      else
        same = 0;
      if (prev == 0)
        ++zero;
      else
        ++one;
    }
    if (same)
      A[i] = 0;
    else if (swap)
      A[i] = 1;
    else if (zero == 2)
      A[i] = 2;
    else
      A[i] = 3;
  }
  for (int i = 0; i < 10; ++i) {
    if (A[i] == 0)
      B[1] |= (1LL << i);
    else if (A[i] == 1)
      B[0] |= (1LL << i), B[1] |= (1LL << i);
    else if (A[i] == 2)
      ;
    else if (A[i] == 3)
      B[1] |= (1LL << i), B[2] |= (1LL << i);
  }
  cout << 3 << endl;
  cout << "^ " << B[0] << endl;
  cout << "& " << B[1] << endl;
  cout << "| " << B[2] << endl;
  return 0;
}

```