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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 0;
int n;
vector<pair<char, int> > v, ans;
int ch[11][2];
bool ok(int fir, int sec) {
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 1; j++) {
      int one = j;
      int two = j;
      one = ch[i][j];
      if (((ans[fir].second) & (1 << i))) {
        if (ans[fir].first == '|') {
          two = 1;
        } else if (ans[fir].first == '^') {
          two = two ^ 1;
        }
      }
      if (((ans[sec].second) & (1 << i))) {
        if (ans[sec].first == '|') {
          two = 1;
        } else if (ans[sec].first == '^') {
          two = two ^ 1;
        }
      }
      if (one != two) return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  char op;
  int num;
  for (int i = 0; i < n; i++) {
    scanf("%c", &op);
    scanf("%c", &op);
    scanf("%d", &num);
    v.push_back(make_pair(op, num));
  }
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 1; j++) {
      int now = j;
      for (int k = 0; k < n; k++) {
        if (((v[k].second) & (1 << i))) {
          if (v[k].first == '|') {
            now = 1;
          } else if (v[k].first == '^') {
            now = now ^ 1;
          }
        } else {
          if (v[k].first == '&') now = 0;
        }
      }
      ch[i][j] = now;
    }
  }
  ans.clear();
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 1024; j++) {
      char oper;
      if (i == 0)
        oper = '^';
      else if (i == 1)
        oper = '|';
      ans.push_back(make_pair(oper, j));
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    for (int j = 0; j < ans.size(); j++) {
      if (ok(i, j)) {
        cout << 2 << endl;
        cout << ans[i].first << " " << ans[i].second << endl;
        cout << ans[j].first << " " << ans[j].second << endl;
        return 0;
      }
    }
  }
}