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 a[15][2], N, op[15], ans[15];
char s[5];
int main() {
  scanf("%d", &N);
  for (int i = 0; i <= 9; i++) a[i][1] = 1;
  while (N--) {
    int v;
    scanf("%s%d", s, &v);
    for (int i = 0; i <= 9; i++)
      for (int j = 0; j <= 1; j++) {
        int o = ((v >> i) & 1);
        if (s[0] == '|') a[i][j] |= o;
        if (s[0] == '&') a[i][j] &= o;
        if (s[0] == '^') a[i][j] ^= o;
      }
  }
  for (int i = 0; i <= 9; i++) {
    if (a[i][0] == 0 && a[i][1] == 0) {
      op[i] = 3;
    }
    if (a[i][0] == 1 && a[i][1] == 0) {
      op[i] = 1;
    }
    if (a[i][0] == 1 && a[i][1] == 1) {
      op[i] = 2;
    }
  }
  ans[1] = 0;
  ans[2] = 0;
  ans[3] = (1 << (10)) - 1;
  for (int i = 0; i <= 9; i++) {
    ans[op[i]] ^= (1 << (i));
  }
  puts("3");
  printf("^ %d\n", ans[1]);
  printf("| %d\n", ans[2]);
  printf("& %d\n", ans[3]);
  return 0;
}