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;
const int inf = 0x3f3f3f3f;
const int maxn = 5e5 + 5;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    int a = 0, b = 1023;
    for (int i = 0; i < n; i++) {
      getchar();
      char ch;
      int x;
      scanf("%c %d", &ch, &x);
      if (ch == '|') {
        a = a | x;
        b = b | x;
      } else if (ch == '&') {
        a = a & x;
        b = b & x;
      } else {
        a = a ^ x;
        b = b ^ x;
      }
    }
    int and1 = a | b, or1 = a & b, xor1 = a & (1023 ^ b);
    cout << 3 << endl;
    cout << "& " << and1 << endl;
    cout << "| " << or1 << endl;
    cout << "^ " << xor1 << endl;
  }
  return 0;
}