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 const Max_N = 1e6 + 5;
int n;
char op[Max_N];
int x[Max_N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    char s[5];
    scanf("%s%d", s, x + i);
    op[i] = s[0];
  }
  int ad = 0, xr = 0, r = 0;
  for (int i = 0; i < 10; ++i) {
    int t0 = 0;
    int t1 = 1;
    for (int j = 0; j < n; ++j) {
      int k = (x[j] >> i) & 1;
      if (op[j] == '|') {
        t0 |= k;
        t1 |= k;
      } else if (op[j] == '&') {
        t0 &= k;
        t1 &= k;
      } else {
        t0 ^= k;
        t1 ^= k;
      }
    }
    if (t0 == 0 && t1 == 1) {
      ad |= (1 << i);
    } else if (t0 == 0 && t1 == 0) {
    } else if (t0 == 1 && t1 == 0) {
      ad |= (1 << i);
      xr |= (1 << i);
    } else if (t0 == 1 && t1 == 1) {
      ad |= (1 << i);
      r |= (1 << i);
    }
  }
  printf("3\n");
  ;
  printf("^ %d\n", xr);
  printf("| %d\n", r);
  printf("& %d\n", ad);
  return 0;
}