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.
#include <bits/stdc++.h>
using namespace std;
typedef struct data {
  long long q;
  long long r;
} data;
long long A[1000];
char c[100];
char s[100];
int main() {
  long long T = 0, N = 0, C = 0, R = 0, K = 0, i = 0, j = 0, len = 0, end = 0,
            d = 0, y = 0, x = 0, z = 0, B = 1023, D = 0, p = 1;
  scanf("%lld", &N);
  for (i = 0; i <= 9; i++) {
    A[i] = 2;
  }
  for (j = 0; j < N; j++) {
    scanf("%s%lld", c, &d);
    for (i = 0; i <= 9; i++) {
      z = d % 2;
      switch (c[0]) {
        case '|':
          if (z == 1) A[i] = 1;
          break;
        case '&':
          if (z == 0) A[i] = 0;
          break;
        case '^':
          if (z == 1) {
            if (A[i] == 0)
              A[i] = 1;
            else if (A[i] == 1)
              A[i] = 0;
            else if (A[i] == 2)
              A[i] = 3;
            else if (A[i] == 3)
              A[i] = 2;
          }
          break;
        default:
          break;
      }
      d = d / 2;
    }
  }
  p = 1;
  for (i = 0; i <= 9; i++) {
    if (A[i] == 0) {
      B = B - p;
    } else if (A[i] == 1) {
      C = C + p;
    } else if (A[i] == 3) {
      D = D + p;
    }
    p = p * 2;
  }
  printf("3\n");
  printf("& %lld\n", B);
  printf("| %lld\n", C);
  printf("^ %lld\n", D);
  return 0;
}
