#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ll inf = 1e18;
void solve() {
  int n;
  cin >> n;
  unsigned int xx = 42;
  unsigned int yy = xx;
  vector<int> state(10);
  bool special = false;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    unsigned int x;
    cin >> x;
    bitset<10> bt(x);
    if (c == '|')
      xx |= x;
    else if (c == '&')
      xx &= x;
    else
      xx ^= x;
    for (int i = 0; i < bt.size(); i++) {
      if (bt[i] == 1) {
        if (c == '|') {
          state[i] = 1;
        } else if (c == '^') {
          if (state[i] == 0)
            state[i] = 3;
          else if (state[i] == 1)
            state[i] = 2;
          else if (state[i] == 2)
            state[i] = 1;
          else if (state[i] == 3)
            state[i] = 0;
        }
      } else {
        if (c == '&') {
          state[i] = 2;
        }
      }
    }
  }
  vector<int> pow2(10, 1);
  for (int i = 0; i < pow2.size(); i++) {
    for (int j = 0; j < i; j++) {
      pow2[i] *= 2;
    }
  }
  vector<pair<char, int> > ans;
  int a1 = 0;
  bool ff = 0;
  for (int i = 0; i < pow2.size(); i++) {
    if (state[i] == 1) {
      ff = 1;
      a1 += pow2[i];
    }
  }
  if (ff) ans.push_back({'|', a1});
  int a2 = 0;
  bool f = 0;
  for (int i = 0; i < 10; i++) {
    if (state[i] == 2) {
      f = 1;
      a2 += 0;
    } else
      a2 += pow2[i];
  }
  if (f) ans.push_back({'&', a2});
  int a3 = 0;
  f = 0;
  for (int i = 0; i < 10; i++) {
    if (state[i] == 3) {
      f = 1;
      a3 += pow2[i];
    }
  }
  if (f) ans.push_back({'^', a3});
  cout << ans.size() << endl;
  for (auto p : ans) {
    char c = p.first;
    unsigned int x = p.second;
    if (c == '|')
      yy |= x;
    else if (c == '&')
      yy &= x;
    else
      yy ^= x;
    cout << p.first << ' ' << p.second << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
using ll = long long;
using ld = long double;
const ll inf = 1e18;
void solve() {
  int n;
  cin >> n;
  unsigned int xx = 42;
  unsigned int yy = xx;
  vector<int> state(10);
  bool special = false;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    unsigned int x;
    cin >> x;
    bitset<10> bt(x);
    if (c == '|')
      xx |= x;
    else if (c == '&')
      xx &= x;
    else
      xx ^= x;
    for (int i = 0; i < bt.size(); i++) {
      if (bt[i] == 1) {
        if (c == '|') {
          state[i] = 1;
        } else if (c == '^') {
          if (state[i] == 0)
            state[i] = 3;
          else if (state[i] == 1)
            state[i] = 2;
          else if (state[i] == 2)
            state[i] = 1;
          else if (state[i] == 3)
            state[i] = 0;
        }
      } else {
        if (c == '&') {
          state[i] = 2;
        }
      }
    }
  }
  vector<int> pow2(10, 1);
  for (int i = 0; i < pow2.size(); i++) {
    for (int j = 0; j < i; j++) {
      pow2[i] *= 2;
    }
  }
  vector<pair<char, int> > ans;
  int a1 = 0;
  bool ff = 0;
  for (int i = 0; i < pow2.size(); i++) {
    if (state[i] == 1) {
      ff = 1;
      a1 += pow2[i];
    }
  }
  if (ff) ans.push_back({'|', a1});
  int a2 = 0;
  bool f = 0;
  for (int i = 0; i < 10; i++) {
    if (state[i] == 2) {
      f = 1;
      a2 += 0;
    } else
      a2 += pow2[i];
  }
  if (f) ans.push_back({'&', a2});
  int a3 = 0;
  f = 0;
  for (int i = 0; i < 10; i++) {
    if (state[i] == 3) {
      f = 1;
      a3 += pow2[i];
    }
  }
  if (f) ans.push_back({'^', a3});
  cout << ans.size() << endl;
  for (auto p : ans) {
    char c = p.first;
    unsigned int x = p.second;
    if (c == '|')
      yy |= x;
    else if (c == '&')
      yy &= x;
    else
      yy ^= x;
    cout << p.first << ' ' << p.second << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```