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;
int F[1024];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  char a;
  int b;
  for (int mask = 0; mask < (1 << 10); mask++) {
    F[mask] = mask;
  }
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    for (int mask = 0; mask < (1 << 10); mask++) {
      if (a == '|')
        F[mask] = (F[mask] | b);
      else if (a == '^')
        F[mask] = (F[mask] ^ b);
      else
        F[mask] = (F[mask] & b);
    }
  }
  for (int i = 0; i < (1 << 10); i++) {
    for (int j = 0; j < (1 << 10); j++) {
      set<int> S;
      for (int mask = 0; mask < (1 << 10); mask++) {
        int val = mask;
        val = (val | i);
        val = (val & j);
        S.insert((val ^ F[mask]));
      }
      if (S.size() == 1) {
        cout << 3 << '\n';
        cout << "| " << i << '\n';
        cout << "& " << j << '\n';
        cout << "^ " << (*S.begin()) << '\n';
        return 0;
      }
    }
  }
  return 0;
}