Rightmost different bit

//Rightmost different bit
Given two numbers M and N. The task is to find the position of rightmost different bit in binary representation of numbers.

Input:
The input line contains T, denoting the number of testcases. Each testcase follows. First line of each testcase contains two space separated integers M and N.

Output:
For each testcase in new line, print the position of rightmost different bit in binary representation of numbers. If both M and N are same then print -1 in this case.

Constraints:
1 <= T <= 100
1 <= M <= 103
1 <= N <= 103

Example:
Input:
2
11 9
52 4

Output:
2
5

Explanation:
Tescase 1: Binary representaion of the given numbers are: 1011 and 1001, 2nd bit from right is different.


import java.util.*;
import java.lang.*;
import java.io.*;

class GFG {
	public static void main (String[] args) throws NumberFormatException, IOException{
		//code
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
        int t = Integer.parseInt(br.readLine()); 
    while (t-- > 0) { 
        // n : size of array 
    	String[] val=br.readLine().split(" ");
        int n=Integer.parseInt(val[0]);
        int sum=Integer.parseInt(val[1]);
    	 if(n==sum) {
        	System.out.println(-1);
        }else {
        int s=n^sum;int count=0;
    	while((s&1)!=1) {
    		s=s>>1;
    		count+=1;
    	}
        if(count>0)
        	System.out.println(count+1);
        else if(s>1)
        	System.out.println(1);
    }}}
}