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;
char s[2];
int main() {
  int n, s0 = 0, s1 = 0, ss = 1023, s2 = 0;
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) {
    int x;
    scanf("%s%d", s, &x);
    if (s[0] == '|')
      s0 &= (ss - x), s1 |= x;
    else if (s[0] == '&')
      x = ss - x, s1 &= (ss - x), s0 |= x;
    else {
      int tmp0 = x & s0, tmp1 = x & s1;
      s0 = s0 - tmp0 + tmp1, s1 = s1 - tmp1 + tmp0;
      s2 ^= x;
    }
  }
  printf("3\n");
  printf("| %d\n", s1);
  s0 = ss - s0;
  printf("& %d\n", s0);
  printf("^ %d\n", s2 & (ss - ((ss - s0) | s1)));
  return 0;
}