#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-5;
const int INF = 1e9;
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
const long long pow(const long long p, const long long q) {
  long long t = 1;
  for (int i = (0); i < (int)(q); ++i) t *= p;
  return t;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<char> arb(10, 3);
  for (int i = (0); i < (int)(n); ++i) {
    char c;
    int t;
    cin >> c >> t;
    if (c == '|') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) > 0) {
          arb[i] = 1;
        }
      }
    }
    if (c == '&') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) == 0) {
          arb[i] = 0;
        }
      }
    }
    if (c == '^') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) > 0) {
          arb[i] ^= 1;
        }
      }
    }
  }
  cout << 3 << endl;
  cout << "& ";
  int t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] > 0) {
      t += (1 << i);
    }
  }
  cout << t << endl;
  cout << "| ";
  t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] == 1) {
      t += (1 << i);
    }
  }
  cout << t << endl;
  cout << "^ ";
  t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] == 2) {
      t += (1 << i);
    }
  }
  cout << t << endl;
}

### Prompt

Generate 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 long double eps = 1e-5;
const int INF = 1e9;
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
const long long pow(const long long p, const long long q) {
  long long t = 1;
  for (int i = (0); i < (int)(q); ++i) t *= p;
  return t;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<char> arb(10, 3);
  for (int i = (0); i < (int)(n); ++i) {
    char c;
    int t;
    cin >> c >> t;
    if (c == '|') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) > 0) {
          arb[i] = 1;
        }
      }
    }
    if (c == '&') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) == 0) {
          arb[i] = 0;
        }
      }
    }
    if (c == '^') {
      for (int i = (0); i < (int)(10); ++i) {
        if ((t & (1 << i)) > 0) {
          arb[i] ^= 1;
        }
      }
    }
  }
  cout << 3 << endl;
  cout << "& ";
  int t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] > 0) {
      t += (1 << i);
    }
  }
  cout << t << endl;
  cout << "| ";
  t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] == 1) {
      t += (1 << i);
    }
  }
  cout << t << endl;
  cout << "^ ";
  t = 0;
  for (int i = (0); i < (int)(10); ++i) {
    if (arb[i] == 2) {
      t += (1 << i);
    }
  }
  cout << t << endl;
}

```