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;
int a[500005];
char ch[500005];
int main() {
  int cases, caseno = 0, n, i, j, k, cnt, sum;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf(" %c %d", &ch[i], &a[i]);
  }
  int x = 0, y = 0, z = 0;
  for (i = 0; i <= 9; i++) {
    int Or = 0, Xor = 0, And = 1;
    for (j = 1; j <= n; j++) {
      if (ch[j] == '|' && (a[j] & (1 << i))) {
        Or = 1;
        And = 1;
        Xor = 0;
      } else if (ch[j] == '&' && !(a[j] & (1 << i))) {
        Or = 0;
        And = 0;
        Xor = 0;
      } else if (ch[j] == '^' && (a[j] & (1 << i))) {
        if (Or == 1) {
          Or = 0;
          Xor = 0;
          And = 0;
        } else if (Or == 0 && And == 1) {
          Xor = (Xor + 1) % 2;
        } else if (And == 0) {
          Or = 1;
          Xor = 0;
          And = 1;
        }
      }
    }
    if (Or) x = x | (1 << i);
    if (And) y = y | (1 << i);
    if (Xor) z = z | (1 << i);
  }
  printf("3\n");
  printf("| %d\n", x);
  printf("& %d\n", y);
  printf("^ %d", z);
  return 0;
}
