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 main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  int A[12];
  int B[12];
  for (int i = 0; i < 10; i++) A[i] = true;
  for (int i = 0; i < 10; i++) B[i] = false;
  for (int i = 0; i < n; i++) {
    char x;
    int c;
    cin >> x >> c;
    for (int j = 0; j < 10; j++) {
      if (x == '&') {
        A[j] &= (c % 2);
        B[j] &= (c % 2);
      }
      if (x == '|') {
        A[j] |= (c % 2);
        B[j] |= (c % 2);
      }
      if (x == '^') {
        A[j] ^= (c % 2);
        B[j] ^= (c % 2);
      }
      c /= 2;
    }
  }
  int AAA = 0;
  int BBB = 0;
  for (int i = 0; i < 10; i++) {
    int a_ = 0, b_ = 0;
    for (int a = 0; a < 2; a++) {
      for (int b = 0; b < 2; b++) {
        int AA = (((true | a) ^ b));
        int BB = (((false | a) ^ b));
        if (AA == A[i] && BB == B[i]) {
          a_ = a;
          b_ = b;
        }
      }
    }
    AAA += a_ * (1ll << i);
    BBB += b_ * (1ll << i);
  }
  cout << 2 << "\n";
  cout << "| " << AAA << '\n';
  cout << "^ " << BBB << '\n';
}
