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>
int main() {
  int n;
  scanf("%d", &n);
  int a = 0, b = 1023;
  for (int i = 1; i <= n; i++) {
    int x;
    char c;
    getchar();
    scanf("%c %d", &c, &x);
    if (c == '|') a |= x, b |= x;
    if (c == '^') a ^= x, b ^= x;
    ;
    if (c == '&') a &= x, b &= x;
  }
  int s1 = 0, s2 = 0, s3 = 1023;
  for (int i = 0; i < 10; i++) {
    int c1 = (a >> i) & 1, c2 = (b >> i) & 1;
    if (c1 == c2)
      if (c1 == 1)
        s1 += (1 << i);
      else
        s3 -= 1 << i;
    else if (c1 == 1)
      s2 += (1 << i);
    else
      ;
  }
  int num = 0;
  if (s1) num++;
  if (s2) num++;
  if (s3 != 1023) num++;
  printf("%d\n", num);
  if (s3 != 1023) printf("%c %d\n", '&', s3);
  if (s2) printf("%c %d\n", '^', s2);
  if (s1) printf("%c %d\n", '|', s1);
}