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 bits[10];
int main() {
  int n, x;
  char c;
  scanf("%d\n", &n);
  memset(bits, -1, sizeof bits);
  for (int i = 0; i < n; i++) {
    scanf("%c %d\n", &c, &x);
    if (c == '&') {
      for (int j = 0; j < 10; j++) {
        if (x & (1 << j)) continue;
        bits[j] = 0;
      }
    } else if (c == '|') {
      for (int j = 0; j < 10; j++) {
        if (x & (1 << j)) {
          bits[j] = 1;
        }
      }
    } else {
      for (int j = 0; j < 10; j++) {
        if (x & (1 << j)) {
          if (bits[j] == 0)
            bits[j] = 1;
          else if (bits[j] == 1)
            bits[j] = 0;
          else if (bits[j] == -1)
            bits[j] = -2;
          else
            bits[j] = -1;
        }
      }
    }
  }
  int and_ans, or_ans, xor_ans;
  and_ans = (1 << 10) - 1;
  or_ans = 0;
  xor_ans = 0;
  for (int i = 0; i < 10; i++) {
    if (bits[i] == 1)
      or_ans |= (1 << i);
    else if (bits[i] == 0)
      and_ans &= ~(1 << i);
    else if (bits[i] == -2)
      xor_ans |= (1 << i);
  }
  printf("3\n");
  printf("& %d\n", and_ans);
  printf("| %d\n", or_ans);
  printf("^ %d\n", xor_ans);
  return 0;
}