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() {
  long long int n;
  cin >> n;
  long long int a[11];
  for (int i = 0; i < 11; ++i) a[i] = 2;
  for (int i = 0; i < n; ++i) {
    char o;
    int x;
    cin >> o >> x;
    if (o == '|') {
      for (int i = 0; i < 11; ++i) {
        if (x % 2 == 1) a[i] = 1;
        x /= 2;
      }
    } else if (o == '^') {
      for (int i = 0; i < 11; ++i) {
        if (x % 2 == 1) {
          if (a[i] == -1)
            a[i] = 2;
          else if (a[i] == 2)
            a[i] = -1;
          else if (a[i] == 0)
            a[i] = 1;
          else
            a[i] = 0;
        }
        x /= 2;
      }
    } else {
      for (int i = 0; i < 11; ++i) {
        if (x % 2 == 0) a[i] = 0;
        x /= 2;
      }
    }
  }
  long long int a1 = 0, a2 = 0, a3 = 0;
  long long int y = 1;
  for (int i = 0; i < 10; ++i) {
    if (a[i] == -1) {
      a2 += y;
      a3 += y;
    } else if (a[i] == 1) {
      a1 += y;
      a3 += y;
    } else if (a[i] == 2)
      a3 += y;
    y *= 2;
  }
  cout << 3 << endl;
  cout << "| " << a1 << endl;
  cout << "^ " << a2 << endl;
  cout << "& " << a3 << endl;
}