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.
#include <bits/stdc++.h>
using namespace std;
signed main() {
  long long n;
  cin >> n;
  char c[n];
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> c[i] >> a[i];
  vector<long long> final;
  final.push_back(0);
  final.push_back(1023);
  final.push_back(0);
  for (long long j = 0; j < 10; j++) {
    long long ans[2];
    long long current = 0;
    for (long long i = 0; i < n; i++) {
      long long bit = ((a[i] & (1LL << j))) != 0;
      if (c[i] == '|')
        current |= bit;
      else if (c[i] == '&')
        current &= bit;
      else if (c[i] == '^')
        current ^= bit;
    }
    ans[0] = current;
    current = 1;
    for (long long i = 0; i < n; i++) {
      long long bit = ((a[i] & (1LL << j)) != 0);
      if (c[i] == '|')
        current |= bit;
      else if (c[i] == '&')
        current &= bit;
      else if (c[i] == '^')
        current ^= bit;
    }
    ans[1] = current;
    if (ans[0] == 0 && ans[1] == 0) {
      final[1] ^= (1LL << j);
    } else if (ans[0] == 0 && ans[1] == 1) {
    } else if (ans[0] == 1 && ans[1] == 0) {
      final[0] ^= (1LL << j);
    } else {
      final[1] ^= (1LL << j);
      final[2] ^= (1LL << j);
    }
  }
  cout << 3 << endl;
  cout << "^ " << final[0] << endl;
  cout << "& " << final[1] << endl;
  cout << "| " << final[2] << endl;
}
