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 mpow = 10;
int main() {
  int n;
  scanf("%i", &n);
  int one[mpow];
  int zero[mpow];
  for (int i = 0; i < mpow; i++) one[i] = 1;
  for (int i = 0; i < mpow; i++) zero[i] = 0;
  for (int i = 0; i < n; i++) {
    char c;
    int x;
    scanf(" %c %i", &c, &x);
    for (int j = 0; j < mpow; j++) {
      int tmp = 0;
      if (x & (1 << j)) tmp = 1;
      if (c == '|') {
        one[j] |= tmp;
        zero[j] |= tmp;
      } else if (c == '^') {
        one[j] ^= tmp;
        zero[j] ^= tmp;
      } else {
        one[j] &= tmp;
        zero[j] &= tmp;
      }
    }
  }
  int or_val = 0;
  int and_val = 0;
  int xor_val = 0;
  for (int i = 0; i < mpow; i++) {
    if (one[i] == 1 and zero[i] == 0) {
      and_val += 1 << i;
    } else if (one[i] == 1 and zero[i] == 1) {
      and_val += 1 << i;
      or_val += 1 << i;
    } else if (one[i] == 0 and zero[i] == 1) {
      and_val += 1 << i;
      xor_val += 1 << i;
    } else if (one[i] == 0 and zero[i] == 0) {
    }
  }
  printf("3 \n");
  printf("| %i\n", or_val);
  printf("& %i\n", and_val);
  printf("^ %i\n", xor_val);
  for (int i = 0; i < mpow; i++) {
    int tmp1 = 1;
    int tmp2 = 0;
    int tmp = 0;
    if (or_val & (1 << i)) tmp = 1;
    tmp1 |= tmp;
    tmp2 |= tmp;
    tmp = 0;
    if (and_val & (1 << i)) tmp = 1;
    tmp1 &= tmp;
    tmp2 &= tmp;
    tmp = 0;
    if (xor_val & (1 << i)) tmp = 1;
    tmp1 ^= tmp;
    tmp2 ^= tmp;
  }
  return 0;
}
