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 = 5e5 + 10;
char op[maxn][2];
int num[maxn];
int main(void) {
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    scanf("%s%d", op[i], &num[i]);
  }
  int and_num = (1 << 10) - 1, or_num = 0, xor_num = 0;
  for (int i = 0; i < 10; i++) {
    int tmp_bit[2];
    tmp_bit[0] = 0, tmp_bit[1] = 1;
    for (int k = 0; k < 2; k++) {
      for (int j = 0; j < N; j++) {
        int cur_bit = (num[j] >> i) & 1;
        if (op[j][0] == '|') {
          tmp_bit[k] |= cur_bit;
        } else if (op[j][0] == '^') {
          tmp_bit[k] ^= cur_bit;
        } else {
          tmp_bit[k] &= cur_bit;
        }
      }
    }
    if (tmp_bit[0] == 0 && tmp_bit[1] == 0) {
      and_num ^= (1 << i);
    } else if (tmp_bit[0] == 0 && tmp_bit[1] == 1) {
    } else if (tmp_bit[0] == 1 && tmp_bit[1] == 0) {
      xor_num |= (1 << i);
    } else {
      or_num |= (1 << i);
    }
  }
  printf("3\n");
  printf("& %d\n", and_num);
  printf("| %d\n", or_num);
  printf("^ %d\n", xor_num);
  return 0;
}
