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;
const int maxn = 5e5 + 10;
char s[maxn];
int a[maxn];
int trans[10][2];
int _xor[10], _or[10], _and[10];
char c;
int main() {
  int n;
  scanf("%d", &n);
  getchar();
  for (int i = 0; i < 10; i++) _and[i] = 1;
  for (int i = 0; i < n; i++) scanf("%c%d%c", &s[i], &a[i], &c);
  for (int i = 0; i < 10; i++) trans[i][0] = 0, trans[i][1] = 1;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 10; j++) {
      for (int k = 0; k < n; k++) {
        if (s[k] == '|')
          trans[j][i] |= ((a[k] >> j) & 1);
        else if (s[k] == '&')
          trans[j][i] &= ((a[k] >> j) & 1);
        else if (s[k] == '^')
          trans[j][i] ^= ((a[k] >> j) & 1);
      }
    }
  }
  for (int i = 0; i < 10; i++) {
    if (trans[i][0] == 1 && trans[i][1] == 1)
      _or[i] = 1;
    else if (trans[i][0] == 0 && trans[i][1] == 0)
      _and[i] = 0;
    else if (trans[i][0] == 1 && trans[i][1] == 0)
      _xor[i] = 1;
  }
  printf("%d\n", 3);
  int num = 0;
  for (int i = 0; i < 10; i++) {
    num += _or[i] ? (1 << i) : 0;
  }
  printf("| %d\n", num);
  num = 0;
  for (int i = 0; i < 10; i++) {
    num += _and[i] ? (1 << i) : 0;
  }
  printf("& %d\n", num);
  num = 0;
  for (int i = 0; i < 10; i++) {
    num += _xor[i] ? (1 << i) : 0;
  }
  printf("^ %d\n", num);
  return 0;
}