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.io.*;
  public class MyClass {
      public static int Xor(int n){
          int count=0;
          if ((n) % 4 == 0)
              count=n;
          else if ((n) % 4 == 1)
              count=1;
          else if ((n) % 4 == 2)
              count=n+1;
          else count=0;
          return count;
      }
      public static int X(int a,int b){
          if(Xor(a-1)==b) return a;
          if((Xor(a-1)^a)==b) return a+2;
          return a+1;
      }
      public static void main(String args[])throws IOException {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          int t=Integer.parseInt(br.readLine());
          while(t-->0){
              String s[]=br.readLine().split(" ");
              int MEX=Integer.parseInt(s[0]);
              int XOR=Integer.parseInt(s[1]);
              System.out.println(X(MEX,XOR));
          }
      }
  }