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;
const int maxn = 5e5 + 10;
char ope[maxn];
int x[maxn];
int n;
int k;
int rx[3];
int one = 1023;
int zero = 0;
void compute(int i) {
  int done = (one >> i) & 1;
  int dzero = (zero >> i) & 1;
  for (int j = 0; j <= 1; j++) {
    for (int k = 0; k <= 1; k++) {
      for (int l = 0; l <= 1; l++) {
        if (((((1 ^ j) & k) | l) == done) && ((((0 ^ j) & k) | l) == dzero)) {
          rx[0] += j << i;
          rx[1] += k << i;
          rx[2] += l << i;
          return;
        }
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    getchar();
    scanf("%c %d", ope + i, x + i);
    if (ope[i] == '^') {
      one ^= x[i];
      zero ^= x[i];
    } else if (ope[i] == '&') {
      one &= x[i];
      zero &= x[i];
    } else {
      one |= x[i];
      zero |= x[i];
    }
  }
  for (int i = 0; 1023 >> i; i++) {
    compute(i);
  }
  printf("%d\n", 3);
  printf("^ %d\n", rx[0]);
  printf("& %d\n", rx[1]);
  printf("| %d\n", rx[2]);
}