description: 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.
solution: import java.util.Arrays;
  import java.util.Scanner;
  
  public class Main
  {
  	public static void main(String[] args)
  	{
  		Scanner sc = new Scanner(System.in);
  		while (sc.hasNext())
  		{
  			int t=sc.nextInt();
  			while(t-->0)
  			{
  				int a=sc.nextInt();
  				int b=sc.nextInt();
  				int c=(a-1)%4;
  				int ans=0;
  			    if (c== 0)ans = a-1;
  		        if (c == 1)ans = 1;
  		        if (c == 2)ans = a;
  		        if (c == 3)ans = 0;
  		        if(ans==b)System.out.println(a);
  		        else 
  		        {
  		        	if((ans^b)==a)
  		        	{
  		        		System.out.println(a+2);
  		        	}
  		        	else System.out.println(a+1);
  		        }
  			}
  		}
  	}
  }