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.lang.Math;
import java.io.* ;


public class Main {
	
	public static void main (String[] args) {
		
		Scanner scan = new Scanner(System.in) ;
		StringBuilder res = new StringBuilder() ;
		int TC= scan.nextInt() ;
		for(int p=0 ;p<TC ;p++) {
			int a= scan.nextInt() ;
			int b= scan.nextInt() ;
			int count= a ;
			int xor = computeXOR(a-1) ;
			xor = xor ^ b ; 
			if(xor == 0) {
				
			}else if(xor == a) {
				count += 2 ;
			}else {
				count++ ;
			}
			res.append(count+"\n") ;
 		}
		System.out.print(res);
	}
	
	public static int computeXOR(int n)
    {
        if (n % 4 == 0)
            return n;
      
        if (n % 4 == 1)
            return 1;
      
        if (n % 4 == 2)
            return n + 1;
      
        return 0;
    }
	
		
}