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.*;
  import java.util.*;
   
  public class Prime {
      static FastReader s=new FastReader();
      static PrintWriter out=new PrintWriter(System.out);
      public static int I(){return s.I();}
      public static long L(){return s.L();}
      public static double D(){return s.D();}
      public static String S(){return s.S();}
      static class FastReader { 
   
          BufferedReader br; 
   
          StringTokenizer st; 
   
   
   
          public FastReader() 
   
          { 
   
              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 I() { return Integer.parseInt(next()); } 
   
   
   
          long L() { return Long.parseLong(next()); } 
   
   
   
          double D() 
   
          { 
   
             return Double.parseDouble(next()); 
   
          } 
   
   
   
          String S() 
   
          { 
   
              String str = ""; 
   
              try { 
   
                 str = br.readLine(); 
   
              } 
   
              catch (IOException e) { 
   
               e.printStackTrace(); 
   
              } 
   
              return str; 
   
          } 
   
      } 
   
      public static void main(String[] args)
      {
          int t=I();
          while(t--!=0){
              int a=I();
              int b=I();
              int xor=0;
              if(a%4==0)xor=0;
              else if(a%4==1)xor=a-1;
              else if(a%4==2)xor=1;
              else if(a%4==3)xor=a;
              
              if(xor==b)System.out.println (a);
              else if((xor^b)!=a)System.out.println (a+1);
              else System.out.println(a+2);
          }
          out.close();
  	}
  }