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.*;
  //3 2 4 2 3
  public class Main{
      public static void main(String[] args){
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while(t-- > 0){
              int a = sc.nextInt();
              int b = sc.nextInt();
              ArrayList<Integer> list = new ArrayList<Integer>();
              // System.out.println()
              if(a == 0){
                  System.out.println("1");
                  continue;
              }
              int xor = computeXOR(a - 1);
              if(xor == b){
                  System.out.println(a);
                  continue;
              }
              if((xor ^ a) == b){
                  System.out.println(a + 2);
                  continue;
              }
              else{
                  System.out.println(a + 1);
                  continue;
              }
              // if(b > a){
              //     if(xor == b){
              //         System.out.println(a);
              //     }
              //     else{
              //         int ans = a + 1;
              //         System.out.println(ans);
              //     }
              // }
              // else{
              //     if(xor == b){
              //         System.out.println(a);
              //     }
              //     else{
              //         int ans = a + 2;
              //         System.out.println(ans);
              //     }
              // }
          }
      }
      public static int computeXOR(int n)
      {
         
        // If n is a multiple of 4
        if (n % 4 == 0)
          return n;
       
        // If n%4 gives remainder 1
        if (n % 4 == 1)
          return 1;
       
        // If n%4 gives remainder 2
        if (n % 4 == 2)
          return n + 1;
       
        // If n%4 gives remainder 3
        return 0;
      }
  }