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;
const int maxn = 20;
int n, c;
char ch[2];
int num[maxn];
int main() {
  ios::sync_with_stdio(false);
  while (scanf("%d", &n) != EOF) {
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; ++i) {
      scanf("%s%d", ch, &c);
      for (int j = 0; j < 10; ++j) {
        int tmp = ((c >> j) & 1);
        if (ch[0] == '&') {
          if (tmp == 0) {
            num[j] = 3;
          }
        } else if (ch[0] == '|') {
          if (tmp == 1) {
            num[j] = 2;
          }
        } else if (ch[0] == '^') {
          if (tmp == 1) {
            num[j] ^= 1;
          }
        }
      }
    }
    int Xor = 0, Or = 0, And = 1023;
    for (int i = 0; i < 10; ++i) {
      if (num[i] == 1) {
        Xor |= (1 << i);
      } else if (num[i] == 2) {
        Or |= (1 << i);
      } else if (num[i] == 3) {
        And ^= (1 << i);
      }
    }
    printf("3\n");
    printf("| %d\n", Or);
    printf("^ %d\n", Xor);
    printf("& %d\n", And);
  }
  return 0;
}
