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() {
  int n;
  scanf("%d\n", &n);
  vector<char> op(n);
  vector<int> val(n);
  int res[10][2];
  for (int i = 0; i < n; ++i) {
    scanf("%c %d\n", &op[i], &val[i]);
  }
  for (int i = 0; i < 10; ++i) {
    for (int j = 0; j < 2; ++j) {
      int sb = j;
      for (int k = 0; k < n; ++k) {
        int b = (val[k] & (1 << i)) > 0 ? 1 : 0;
        if (op[k] == '|') {
          sb |= b;
        } else if (op[k] == '&') {
          sb &= b;
        } else if (op[k] == '^') {
          sb ^= b;
        }
      }
      res[i][j] = sb;
    }
  }
  int v_and = (1 << 10) - 1;
  int v_or = 0;
  int v_xor = 0;
  for (int i = 0; i < 10; ++i) {
    int type = res[i][0] * 2 + res[i][1];
    if (type == 0) {
      v_and -= (1 << i);
    } else if (type == 1) {
    } else if (type == 2) {
      v_xor += (1 << i);
    } else if (type == 3) {
      v_or += (1 << i);
    }
  }
  printf("3\n");
  printf("& %d\n", v_and);
  printf("| %d\n", v_or);
  printf("^ %d\n", v_xor);
}