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 main() {
  int n, x;
  char op[3];
  int entry[10];
  for (int i = 0; i < 10; ++i) entry[i] = 3;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s %d", op, &x);
    if (op[0] == '&')
      for (int j = 0; j < 10; ++j) {
        if (!(x & (1 << j))) entry[j] = 0;
      }
    else if (op[0] == '|')
      for (int j = 0; j < 10; ++j) {
        if (x & (1 << j)) entry[j] = 1;
      }
    else if (op[0] == '^')
      for (int j = 0; j < 10; ++j) {
        if (entry[j] == 3 && (x & (1 << j)))
          entry[j] = 2;
        else if (entry[j] == 2 && (x & (1 << j)))
          entry[j] = 3;
        else if (entry[j] == 1 && (x & (1 << j)))
          entry[j] = 0;
        else if (entry[j] == 0 && (x & (1 << j)))
          entry[j] = 1;
      }
  }
  int AND = (1 << 10) - 1, XOR = 0, OR = 0;
  for (int i = 0; i < 10; ++i) {
    if (entry[i] == 0)
      AND &= (1023 & ~(1 << i));
    else if (entry[i] == 1)
      OR |= (1 << i);
    else if (entry[i] == 2)
      XOR |= (1 << i);
  }
  printf("3\n");
  printf("& %d\n", AND);
  printf("| %d\n", OR);
  printf("^ %d\n", XOR);
  return 0;
}