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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class A {
      public static void main(String[]args){
          MyScanner sc = new MyScanner();
          int t = sc.nextInt();
          for(int i=0;i<t;i++){
              int a = sc.nextInt();
              int b = sc.nextInt();
              int max=a-1;
              int count=0;
              if(max%4==0){
                  count=max;
              }
              else if(max%4==1){
                  count=1;
              }
              else if(max%4==2){
                  count=a;
              }
              else{
                  count=0;
              }
  
              if(count==b){
                  System.out.println(a);
              }
              else if((count^a)==b){System.out.println(a+2);}
              else{
                  System.out.println(a+1);
                  
              }
              
          }
  
          
      }
  
  }
  class MyScanner {
      BufferedReader br;
      StringTokenizer st;
  
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
  
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int nextInt() {
          return Integer.parseInt(next());
      }
  
      long nextLong() {
          return Long.parseLong(next());
      }
  
      double nextDouble() {
          return Double.parseDouble(next());
      }
  
      String nextLine(){
          String str = "";
      try {
         str = br.readLine();
      } catch (IOException e) {
         e.printStackTrace();
      }
      return str;
      }
  
   }