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;
const int maxn = 520;
int a[20], b[20];
int main() {
  int n, c, i, j, a1, a2, a3;
  char ch;
  scanf("%d", &n);
  while (n--) {
    scanf(" %c %d", &ch, &c);
    if (ch == '|') {
      for (i = 10; i >= 0; i--) {
        if (((c >> i) & 1) == 1) b[i] = 1, a[i] = 1;
      }
    } else if (ch == '&') {
      for (i = 10; i >= 0; i--) {
        if (((c >> i) & 1) == 0) b[i] = 1, a[i] = 0;
      }
    } else if (ch == '^') {
      for (i = 10; i >= 0; i--) {
        if (((c >> i) & 1) == 1) {
          if (b[i] == 0)
            b[i] = 2, a[i] ^= 1;
          else
            a[i] ^= 1;
        }
      }
    }
  }
  a1 = a2 = a3 = 0;
  for (i = 9; i >= 0; i--) {
    if (b[i] == 2) {
      a1 = a1 | (a[i] << i);
      a2 = a2 | (0 << i);
      a3 = a3 | (1 << i);
    } else if (b[i] == 1) {
      if (a[i] == 1) a2 = a2 | (1 << i), a3 = a3 | (1 << i);
    } else {
      a3 = a3 | (1 << i);
    }
  }
  printf("3\n^ %d\n& %d\n| %d\n", a1, a3, a2);
  return 0;
}
