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 res[13][2];
char tmp[10];
int main() {
  int n;
  scanf("%d", &n);
  for (int j = 0; j < 10; j++) res[j][0] = 0, res[j][1] = 1;
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%s%d", tmp, &v);
    for (int j = 0; j < 10; j++) {
      if (tmp[0] == '|') {
        res[j][0] = (res[j][0] | (v >> j)) & 1;
        res[j][1] = (res[j][1] | (v >> j)) & 1;
      }
      if (tmp[0] == '^') {
        res[j][0] = (res[j][0] ^ (v >> j)) & 1;
        res[j][1] = (res[j][1] ^ (v >> j)) & 1;
      }
      if (tmp[0] == '&') {
        res[j][0] = (res[j][0] & (v >> j)) & 1;
        res[j][1] = (res[j][1] & (v >> j)) & 1;
      }
    }
  }
  int v1 = 0, v2 = 0, v3 = 0;
  for (int j = 0; j < 10; j++) {
    if (res[j][0] == 0 && res[j][1] == 0)
      v1 = v1 | (0 << j);
    else
      v1 = v1 | (1 << j);
    if (res[j][0] == 1 && res[j][1] == 0)
      v2 = v2 | (1 << j);
    else
      v2 = v2 | (0 << j);
    if (res[j][0] == 1 && res[j][1] == 1)
      v3 = v3 | (1 << j);
    else
      v3 = v3 | (0 << j);
  }
  printf("3\n");
  printf("& %d\n", v1);
  printf("^ %d\n", v2);
  printf("| %d\n", v3);
  return 0;
}