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;
void two(vector<int> &arg, int x) {
  int pos = 9;
  while (x > 0) {
    arg[pos] = x % 2;
    pos--;
    x /= 2;
  }
  while (pos >= 0) {
    arg[pos] = 0;
    pos--;
  }
  return;
}
int main() {
  vector<int> num(10, 0), arg(10), sto_proc(10, 0);
  int n;
  cin >> n;
  for (int o = 0; o < n; o++) {
    char t;
    cin >> t;
    int x;
    cin >> x;
    two(arg, x);
    if (t == '|') {
      for (int i = 0; i < 10; i++) {
        if (arg[i]) {
          num[i] = 1;
          sto_proc[i] = 1;
        }
      }
    }
    if (t == '&') {
      for (int i = 0; i < 10; i++) {
        if (arg[i] == 0) {
          num[i] = 0;
          sto_proc[i] = 1;
        }
      }
    }
    if (t == '^') {
      for (int i = 0; i < 10; i++) {
        if (arg[i] == num[i]) {
          num[i] = 0;
        } else {
          num[i] = 1;
        }
      }
    }
  }
  vector<int> and_arg(10, 1), xor_arg(10, 0), or_arg(10, 0);
  for (int i = 0; i < 10; i++) {
    if (num[i] && sto_proc[i]) {
      or_arg[i] = 1;
    }
    if (!num[i] && sto_proc[i]) {
      and_arg[i] = 0;
    }
    if (num[i] && !sto_proc[i]) {
      xor_arg[i] = 1;
    }
  }
  int one = 0, two = 0, three = 0;
  int p = 512;
  for (int i = 0; i < 10; i++) {
    if (and_arg[i]) {
      one += p;
    }
    if (or_arg[i]) {
      two += p;
    }
    if (xor_arg[i]) {
      three += p;
    }
    p /= 2;
  }
  cout << 3 << "\n& " << one << "\n| " << two << "\n^ " << three;
  return 0;
}