Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.

What is the shortest possible length of the array Bob wrote?

Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5; 0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.

Output

For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.

Example

Input


5
1 1
2 1
2 0
1 10000
2 10000


Output


3
2
3
2
3

Note

In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].

In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].

It can be shown that these arrays are the shortest arrays possible.
import java.util.*;
import java.math.*;
public class test

{
  public static void main(String[] args) {
        
        Scanner sc=new Scanner(System.in);
       int t = sc.nextInt();
       for(int l =0;l<t;l++)
       {
    	   int a = sc.nextInt();
    	   int b = sc.nextInt();
    	   int ls = a-1;
    	   int m = a+1;
    	   int max = Integer.MAX_VALUE-1;
    	   int i =0;
    	   int c =0;
    	   if(Math.abs(b-a)%2==1)
	    { //System.out.println(i+" * "+ls);
			   System.out.println(2);
			   c++;
       }
			   
		   
    	   
//    	   while(i<max)
//    	   {
//    		   if((i^ls)==b&&i!=a)
//    		   {  System.out.println(i+" * "+ls);
//    			   System.out.println(2);
//    			   c++;
//    			   break;
//    		   }
//    		   
//    		   i++;
//    		   
//    	   }
    	   if(c==0)
    		   System.out.println(3);
    }
    	   
       }
		 
		
		 
}
	
	
