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.
#include <bits/stdc++.h>
using namespace std;
int arr[1000];
int op(char c, int act, int b) {
  if (c == '|') return act | b;
  if (c == '&') return act & b;
  if (c == '^') return act ^ b;
}
int main() {
  int n, num;
  char c;
  cin >> n;
  int cero = 0, unos = -1;
  for (int i = 0; i < n; i++) {
    cin >> c >> num;
    cero = op(c, cero, num);
    unos = op(c, unos, num);
  }
  int n_or = 0, n_xor = 0;
  for (int i = 0; i < 10; i++) {
    bool pc = cero & (1 << i);
    bool pu = unos & (1 << i);
    if (pc && pu)
      n_or = n_or | (1 << i);
    else if (pc)
      n_xor = n_xor | (1 << i);
    else if (pu)
      ;
    else
      n_or |= (1 << i), n_xor |= (1 << i);
  }
  if (n_xor && n_or)
    printf("2\n| %d\n^ %d\n", n_or, n_xor);
  else if (n_or)
    printf("1\n| %d\n", n_or);
  else if (n_xor)
    printf("1\n^ %d\n", n_xor);
  else
    printf("0\n");
  return 0;
}
