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 f[1005];
char a[1005];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int z, zz;
    for (int i = 0; i <= 10; ++i) f[i] = -1;
    for (int j = 1; j <= n; ++j) {
      scanf("%s", a);
      scanf("%d", &z);
      for (int i = 0; i <= 9; ++i) {
        zz = ((z >> i) & 1);
        if (a[0] == '&') {
          if (!zz) f[i] = 0;
        } else if (a[0] == '|') {
          if (zz) f[i] = 1;
        } else {
          if (zz) {
            if (f[i] == -1)
              f[i] = 2;
            else if (f[i] == 2)
              f[i] = -1;
            else
              f[i] = f[i] ^ 1;
          }
        }
      }
    }
    printf("3\n");
    int aa = 0, bb = 0, cc = 0;
    for (int i = 0; i <= 9; ++i) {
      aa = aa + (1 << i);
      if (f[i] == 0)
        aa = aa - (1 << i);
      else if (f[i] == 1)
        bb = bb + (1 << i);
      else if (f[i] == 2)
        cc = cc + (1 << i);
    }
    printf("& %d\n", aa);
    printf("| %d\n", bb);
    printf("^ %d\n", cc);
  }
}