Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
void clean() {}
void solve() {
  clean();
  int a = 0, b = 1023;
  for (int x, i = 1; i <= n; i++) {
    char s[5];
    scanf("%s %d", s, &x);
    switch (s[0]) {
      case '^':
        a ^= x, b ^= x;
        break;
      case '|':
        a |= x, b |= x;
        break;
      case '&':
        a &= x, b &= x;
        break;
    }
  }
  int A = 1023, X = 0, O = 0, bs = 1;
  for (int i = 0; i < 10; i++) {
    int taki = a & 1, mitsuha = b & 1;
    if (taki == 0) {
      if (mitsuha == 0) A -= bs;
      if (mitsuha == 1)
        ;
    } else if (taki == 1) {
      if (mitsuha == 0) X += bs;
      if (mitsuha == 1) O += bs;
    }
    a >>= 1, b >>= 1, bs <<= 1;
  }
  printf("3\n| %d\n^ %d\n& %d\n", O, X, A);
}
int main() {
  scanf("%d", &n), solve();
  return 0;
}