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.
#include <bits/stdc++.h>
using namespace std;
int getbit(int x, int y) { return (x >> (y - 1)) & 1; }
char c;
int n, x, a[20], _and, _xor, _or, k;
int main() {
  scanf("%d\n", &n);
  for (int i = 1; i <= 10; i++) a[i] = 2;
  for (int i = 1; i <= n; i++) {
    scanf("%c %d\n", &c, &x);
    for (int j = 1; j <= 10; j++) {
      k = getbit(x, j);
      if (c == '|' && k == 1) a[j] = 1;
      if (c == '&' && k == 0) a[j] = 0;
      if (c == '^' && k == 1) {
        if (a[j] == 1)
          a[j] = 0;
        else if (a[j] == 0)
          a[j] = 1;
        else
          a[j] = -a[j];
      }
    }
  }
  _and = 1023;
  for (int i = 1; i <= 10; i++) {
    if (a[i] == -2) _xor = _xor ^ (1 << (i - 1));
    if (a[i] == 0) _and = _and ^ (1 << (i - 1));
    if (a[i] == 1) _or = _or ^ (1 << (i - 1));
  }
  printf("3\n");
  printf("& %d\n", _and);
  printf("| %d\n", _or);
  printf("^ %d\n", _xor);
  return 0;
}
