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;
int n, x[500005];
char c[500005];
int a[20];
int q, w, e;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < 10; i++) a[i] = 2;
  for (int i = 0; i < n; i++) {
    cin >> c[i] >> x[i];
  }
  for (int i = 0; i < 10; i++) {
    int k = (1 << i);
    for (int j = 0; j < n; j++) {
      if (c[j] == '|') {
        if (x[j] & k) {
          a[i] = 1;
        }
      } else if (c[j] == '&') {
        if (!(x[j] & k)) {
          a[i] = 0;
        }
      } else {
        if (x[j] & k) {
          if (a[i] == 0)
            a[i] = 1;
          else if (a[i] == 1)
            a[i] = 0;
          else if (a[i] == 2)
            a[i] = 3;
          else
            a[i] = 2;
        }
      }
    }
  }
  for (int i = 0; i < 10; i++) {
    if (a[i] != 0) q += (1 << i);
    if (a[i] == 3) w += (1 << i);
    if (a[i] == 1) e += (1 << i);
  }
  cout << "3\n";
  cout << "& " << q << "\n";
  cout << "^ " << w << "\n";
  cout << "| " << e << "\n";
  return 0;
}