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 Demo{
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static StringTokenizer st = null;
      static int mod = 1000_000_007;
  
      public static void main(String[] args) throws IOException{
          int testcase = Integer.parseInt(br.readLine());
          int[] xors = new int[300001];
          int temp = 0;
          for(int i=0;i<300001;i++){
              temp^=i;
              xors[i] = temp;
          }
          while(testcase-->0){
              st = new StringTokenizer(br.readLine());
              int n = Integer.parseInt(st.nextToken());
              int m = Integer.parseInt(st.nextToken());
              int t = xors[n-1];
              int ans = n;
  
              if(t!=m){
                  if(t==0 && n==m)ans+=2;
                  else if((t^m)==n){
                      ans+=2;
                  }else{
                      ans++;
                  }
              }
              
              System.out.println(ans);
              
          }
  
      }
  }
  /*
  
  */