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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int i, j, n;
  char ch;
  int num;
  cin >> n;
  int fl[10];
  for (i = 0; i <= 9; i++) fl[i] = 2;
  for (i = 1; i <= n; i++) {
    cin >> ch >> num;
    if (ch == '|') {
      for (j = 0; j <= 9; j++) {
        int z = num & (1 << j);
        if (z > 0) fl[j] = 1;
      }
    } else if (ch == '&') {
      for (j = 0; j <= 9; j++) {
        int z = num & (1 << j);
        if (z == 0) fl[j] = 0;
      }
    } else {
      for (j = 0; j <= 9; j++) {
        int z = num & (1 << j);
        if (z > 0) {
          if (fl[j] == 2)
            fl[j] = 3;
          else if (fl[j] == 3)
            fl[j] = 2;
          else {
            fl[j] = 1 - fl[j];
          }
        }
      }
    }
  }
  int num1 = 0, num2 = 0, num3 = 0;
  for (j = 0; j <= 9; j++) {
    if (fl[j] == 2) {
      num2 |= (1 << j);
    } else if (fl[j] == 3) {
      num2 |= (1 << j);
      num3 |= (1 << j);
    } else {
      if (fl[j] == 1) num2 |= (1 << j);
      num1 |= (1 << j);
    }
  }
  cout << 3 << ('\n');
  cout << "| " << num1 << ('\n');
  cout << "& " << num2 << ('\n');
  cout << "^ " << num3 << ('\n');
  return 0;
}