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 ope[500000 + 10];
char ch[500000 + 10][2];
int main(void) {
  int n;
  int a = 0, b = 0, c = 1023;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d", ch[i], &ope[i]);
  }
  int tar = 1023;
  int tar2 = 0;
  for (int j = 0; j < n; ++j) {
    if (ch[j][0] == '|') {
      tar |= ope[j];
      tar2 |= ope[j];
    } else if (ch[j][0] == '&') {
      tar &= ope[j];
      tar2 &= ope[j];
    } else {
      tar ^= ope[j];
      tar2 ^= ope[j];
    }
  }
  for (int i = 0; i < 10; ++i) {
    if ((tar & (1 << i)) == 0) {
      if ((tar2 & (1 << i)) == 0) {
        c -= (1 << i);
      } else {
        a += (1 << i);
      }
    } else {
      if ((tar2 & (1 << i)) != 0) b += (1 << i);
    }
  }
  printf("3\n^ %d\n| %d\n& %d\n", a, b, c);
  return 0;
}