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 JAVA):

Solution:
import java.io.IOException;
import java.util.Scanner;

public class problem3 {
	public static void main (String [] args) throws IOException {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int one = 1023;
		int zero = 0;
		for(int i = 0; i<n; i++){
			char c = scan.next().charAt(0);
			int x = scan.nextInt(); 
			if(c == '^'){
				one = one^x;
				zero = zero^x;
			}
			else if(c == '&'){
				one = one&x;
				zero = zero&x;
			}

			else{
				one = one|x;
				zero = zero|x;
			}
		}
		System.out.println(3);
		
		
		System.out.println("^ " + (zero&(1023^one)));
		System.out.println("& " + (zero|one));
		System.out.println("| " + (zero&one));
		

	}

}