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>
const int N = 15;
int n, bin[N], f[N][2];
int main() {
  scanf("%d", &n);
  bin[0] = 1;
  for (int i = 1; i <= 10; i++) bin[i] = bin[i - 1] * 2;
  for (int i = 0; i < 10; i++) f[i][1] = 1;
  for (int i = 1; i <= n; i++) {
    char ch[2];
    int x;
    scanf("%s%d", ch, &x);
    if (ch[0] == '|') {
      for (int j = 0; j < 10; j++)
        if (x & bin[j]) f[j][0] = f[j][1] = 1;
    } else if (ch[0] == '&') {
      for (int j = 0; j < 10; j++)
        if (!(x & bin[j])) f[j][0] = f[j][1] = 0;
    } else {
      for (int j = 0; j < 10; j++)
        if (x & bin[j]) f[j][0] ^= 1, f[j][1] ^= 1;
    }
  }
  printf("%d\n", 3);
  int ans = bin[10] - 1;
  for (int i = 0; i < 10; i++)
    if (f[i][0] == 0 && f[i][1] == 0) ans -= bin[i];
  printf("& %d\n", ans);
  ans = 0;
  for (int i = 0; i < 10; i++)
    if (f[i][0] && f[i][1]) ans += bin[i];
  printf("| %d\n", ans);
  ans = 0;
  for (int i = 0; i < 10; i++)
    if (f[i][0] && !f[i][1]) ans += bin[i];
  printf("^ %d\n", ans);
  return 0;
}