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;
const int MAXN = 30;
int bits[MAXN];
int main() {
  fill(bits, bits + MAXN, 3);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    char c;
    int k;
    scanf(" %c %d", &c, &k);
    for (int j = 0; j < (MAXN); j++) {
      int cb = k & (1 << j);
      if (c == '&') {
        if (cb == 0) {
          bits[j] = 0;
        }
      } else if (c == '|') {
        if (cb) {
          bits[j] = 1;
        }
      } else if (c == '^' && cb) {
        if (bits[j] == 2) {
          bits[j] = 3;
        } else if (bits[j] == 3) {
          bits[j] = 2;
        } else if (bits[j] == 0) {
          bits[j] = 1;
        } else if (bits[j] == 1) {
          bits[j] = 0;
        }
      }
    }
  }
  int xv = 0;
  int ov = 0;
  int av = 0b1111111111;
  for (int i = 0; i < (MAXN); i++) {
    int cb = i;
    if (bits[cb] == 1) {
      ov |= 1 << cb;
    }
    if (bits[cb] == 0) {
      av &= ~(1 << cb);
    }
    if (bits[cb] == 2) {
      xv |= (1 << cb);
    }
  }
  printf("3\n");
  printf("^ %d\n", xv);
  printf("| %d\n", ov);
  printf("& %d\n", av);
}