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.*;
import java.lang.*;
import java.util.Scanner;
public class Main {
	public static void main(String[] args){
		Scanner in=new Scanner(System.in);
		int n;n=in.nextInt();
		int[] P=new int[12];
		for(int i=0;i<P.length;i++)P[i]=-1;
		for(int i=0;i<n;i++){
			String s=in.next();
			char c=s.charAt(0);
			int k=in.nextInt();
			for(int j=0;j<10;j++){
				int ins=k%2;
				if(c=='|'){
						if(ins==1)P[j]=1;
						if(k==0)break;
				}
				else if(c=='&'){
					if(ins==0)P[j]=0;
				}
				else if(ins==1){
					if(P[j]==0)P[j]=1;
					else if(P[j]==1)P[j]=0;
					else if(P[j]==-1)P[j]=2;
					else P[j]++;
				}
				k/=2;
			}
			
		}
		int a=0,b=0,c=0;
		for(int i=0;i<10;i++){
			if(P[i]==-1)continue;
			else if(P[i]==0)a|=(1<<i);
			else if(P[i]==1)b|=(1<<i);
			else if(P[i]%2==0)c|=(1<<i);
		}
		//System.out.println(a);
		a=(1<<10)-1-a;
		int ans=0;
		if(a<1023)ans++;
		if(b>0)ans++;
		if(c>0)ans++;
		System.out.println(ans);
		if(a<1023)System.out.println("& "+a);
		if(b>0)System.out.println("| "+b);
		if(c>0)System.out.println("^ "+c);
		
		
	}
}