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() {
  int n;
  cin >> n;
  int biti0 = 0, biti1 = (1 << 10) - 1;
  for (int i = 0; i < n; ++i) {
    char op;
    int x;
    cin >> op >> x;
    if (op == '|') {
      biti0 |= x;
      biti1 |= x;
    } else if (op == '&') {
      biti0 &= x;
      biti1 &= x;
    } else {
      biti0 ^= x;
      biti1 ^= x;
    }
  }
  int ormask = 0, xormask = 0;
  for (int b = 0; b < 10; ++b) {
    int r0 = biti0 & (1 << b), r1 = biti1 & (1 << b);
    if (r0 && r1)
      xormask += (1 << b);
    else if ((!r0) && r1)
      ormask += (1 << b);
    else if (r0 && (!r1)) {
      ormask += (1 << b);
      xormask += (1 << b);
    }
  }
  cout << "2\n"
       << "& " << ormask << "\n^ " << xormask << "\n";
  return 0;
}
