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 N = 510000, B = 10;
int n;
int bits0[B], bits1[B];
int operation(int b, int vb, char c) {
  if (c == '|') {
    return b | vb;
  } else if (c == '^') {
    return b ^ vb;
  } else if (c == '&') {
    return b & vb;
  }
  assert(0);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (B); ++i) {
    bits0[i] = 0;
    bits1[i] = 1;
  }
  for (int i = 0; i < (n); ++i) {
    char c;
    int val;
    scanf(" %c%d", &c, &val);
    for (int i = 0; i < (B); ++i) {
      int vb = !!(val & 1 << i);
      bits0[i] = operation(bits0[i], vb, c);
      bits1[i] = operation(bits1[i], vb, c);
    }
  }
  int valor = 0, valxor = 0, valand = (1 << B) - 1;
  for (int i = 0; i < (B); ++i) {
    int b = 1 << i;
    if (bits0[i] == 1 && bits1[i] == 1) {
      valor |= b;
    } else if (bits0[i] == 0 && bits1[i] == 0) {
      valand &= ~b;
    } else if (bits0[i] == 1 && bits1[i] == 0) {
      valxor |= b;
    }
  }
  printf("3\n");
  printf("| %d\n", valor);
  printf("^ %d\n", valxor);
  printf("& %d\n", valand);
}
