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 main() {
  cin.tie(NULL);
  ios::sync_with_stdio(false);
  ;
  int n, i;
  cin >> n;
  int t1 = 0, t2 = 1023;
  for (i = 0; i < n; i++) {
    char op;
    int con;
    cin >> op;
    cin >> con;
    if (op == '&') {
      t1 &= con;
      t2 &= con;
    }
    if (op == '|') {
      t1 |= con;
      t2 |= con;
    }
    if (op == '^') {
      t1 ^= con;
      t2 ^= con;
    }
  }
  if (t1 == 0 && t2 == 1023) {
    cout << 0 << endl;
  } else {
    int an = 1023, o = 0, x = 0;
    for (i = 0; i < 10; i++) {
      int b1 = ((t1 >> i) & 1);
      int b2 = ((t2 >> i) & 1);
      if (b1 == 0 && b2 == 0) {
        an -= ((1 << i));
      }
      if (b1 == 1 && b2 == 1) {
        o += (1 << i);
      }
      if (b1 == 1 && b2 == 0) {
        x += (1 << i);
      }
    }
    cout << 3 << endl;
    cout << "& " << an << endl;
    cout << "| " << o << endl;
    cout << "^ " << x << endl;
  }
  return 0;
}