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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  short x[10];
  for (int i = 0; i < 10; i++) x[i] = 2;
  for (int i = 0; i < n; i++) {
    char op;
    int a;
    cin >> op >> a;
    bitset<10> b(a);
    if (op == '&') {
      for (int j = 0; j < 10; j++) {
        if (b[9 - j] == 0) x[j] = 0;
      }
    } else if (op == '|') {
      for (int j = 0; j < 10; j++) {
        if (b[9 - j] == 1) x[j] = 1;
      }
    } else {
      for (int j = 0; j < 10; j++) {
        if (b[9 - j] == 1) {
          switch (x[j]) {
            case 0:
              x[j] = 1;
              break;
            case 1:
              x[j] = 0;
              break;
            case 2:
              x[j] = 3;
              break;
            case 3:
              x[j] = 2;
              break;
            default:
              break;
          }
        }
      }
    }
  }
  int andd = 0, orr = 0, xorr = 0;
  for (int i = 0; i < 10; i++) {
    if (x[i] != 0) andd += pow(2, 9 - i);
  }
  for (int i = 0; i < 10; i++) {
    if (x[i] == 1) orr += pow(2, 9 - i);
  }
  for (int i = 0; i < 10; i++) {
    if (x[i] == 3) xorr += pow(2, 9 - i);
  }
  cout << 3 << endl;
  cout << "& " << andd << endl << "| " << orr << endl << "^ " << xorr;
  return 0;
}