#include <bits/stdc++.h>
using namespace std;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
}
struct debug {
  template <class T>
  debug& operator<<(const T&) {
    return *this;
  }
};
using ll = long long;
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void max_self(T& a, T b) {
  a = max(a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<string, int>> ops(n);
  for (int i = 0; i < int(n); i++) {
    cin >> ops[i].first >> ops[i].second;
  }
  vector<int> output;
  for (int start : {0, 1023}) {
    int ans = start;
    for (int i = 0; i < int(n); i++) {
      string s = ops[i].first;
      int y = ops[i].second;
      if (s == "|") {
        ans |= y;
      } else if (s == "&") {
        ans &= y;
      } else {
        ans ^= y;
      }
    }
    output.push_back(ans);
  }
  debug() << "[ "
          << "output"
             ": "
          << (output) << " ] ";
  cout << 2 << endl;
  int a = 0, b = 0;
  for (int bit = 0; (1 << bit) < 1024; bit++) {
    if ((output[0] >> bit & 1) == 1 && (output[1] >> bit & 1) == 1) {
      a |= 1 << bit;
    } else if ((output[0] >> bit & 1) == 0 && (output[1] >> bit & 1) == 0) {
      a |= 1 << bit, b |= 1 << bit;
    } else if ((output[0] >> bit & 1) == 1 && (output[1] >> bit & 1) == 0) {
      b |= 1 << bit;
    } else {
    }
  }
  cout << "| " << a << endl << "^ " << b << endl;
}

### Prompt

Create a solution in cpp for 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;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
}
struct debug {
  template <class T>
  debug& operator<<(const T&) {
    return *this;
  }
};
using ll = long long;
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void max_self(T& a, T b) {
  a = max(a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<string, int>> ops(n);
  for (int i = 0; i < int(n); i++) {
    cin >> ops[i].first >> ops[i].second;
  }
  vector<int> output;
  for (int start : {0, 1023}) {
    int ans = start;
    for (int i = 0; i < int(n); i++) {
      string s = ops[i].first;
      int y = ops[i].second;
      if (s == "|") {
        ans |= y;
      } else if (s == "&") {
        ans &= y;
      } else {
        ans ^= y;
      }
    }
    output.push_back(ans);
  }
  debug() << "[ "
          << "output"
             ": "
          << (output) << " ] ";
  cout << 2 << endl;
  int a = 0, b = 0;
  for (int bit = 0; (1 << bit) < 1024; bit++) {
    if ((output[0] >> bit & 1) == 1 && (output[1] >> bit & 1) == 1) {
      a |= 1 << bit;
    } else if ((output[0] >> bit & 1) == 0 && (output[1] >> bit & 1) == 0) {
      a |= 1 << bit, b |= 1 << bit;
    } else if ((output[0] >> bit & 1) == 1 && (output[1] >> bit & 1) == 0) {
      b |= 1 << bit;
    } else {
    }
  }
  cout << "| " << a << endl << "^ " << b << endl;
}

```