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;
bool res[15][2];
char commands[500005];
int num[500005];
int Xor, Or, And;
int main() {
  int n, i, j, st = 1, j2;
  bool b, b2, BB;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> commands[i] >> num[i];
  }
  for (i = 0; i <= 9; i++) {
    for (j2 = 0; j2 <= 1; j2++) {
      b = j2;
      for (j = 0; j < n; j++) {
        b2 = (num[j] >> i) & 1;
        if (commands[j] == '|') {
          b = b2 | b;
        } else {
          if (commands[j] == '&') {
            b = b2 & b;
          } else {
            b = b2 ^ b;
          }
        }
      }
      if (j2 == 0)
        BB = b;
      else {
        if (BB == 0 && b == 0) {
          Or += st;
          Xor += st;
        }
        if (BB == 1 && b == 0) {
          Xor += st;
        }
        if (BB == 1 && b == 1) {
          Or += st;
        }
      }
    }
    st *= 2;
  }
  cout << "2" << endl;
  cout << "| " << Or << endl;
  cout << "^ " << Xor << endl;
  return 0;
}