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 n;
int a[15];
void f(int fg, int x) {
  if (fg == 1) {
    for (int i = 0; i <= 10; i++) {
      if (((x >> i) & 1) != 0) {
        a[i] = 1;
      }
    }
  } else if (fg == 2) {
    for (int i = 0; i <= 10; i++) {
      if (((x >> i) & 1) == 0) {
        a[i] = 0;
      }
    }
  } else {
    for (int i = 0; i <= 10; i++) {
      if (((x >> i) & 1) != 0) {
        if (a[i] == 0 || a[i] == 1)
          a[i] ^= 1;
        else
          a[i] = -a[i];
      }
    }
  }
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i <= 10; i++) a[i] = 5;
    for (int i = 0; i < n; i++) {
      char ch;
      int tp;
      getchar();
      scanf("%c", &ch);
      scanf("%d", &tp);
      if (ch == '|')
        f(1, tp);
      else if (ch == '&')
        f(2, tp);
      else
        f(3, tp);
    }
    int ansor = 0;
    int ansand = 1023;
    int ansxor = 0;
    for (int i = 0; i < 10; i++) {
      if (a[i] == 1) {
        ansor += (1 << i);
      } else if (a[i] == 0) {
        ansand -= (1 << i);
      } else if (a[i] == -5) {
        ansxor += (1 << i);
      }
    }
    printf("3\n");
    printf("| %d\n", ansor);
    printf("& %d\n", ansand);
    printf("^ %d\n", ansxor);
  }
}