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: //package Contest.CF742;
  
  import java.io.*;
  import java.util.*;
  
  public class B {
      public static void main(String[] args) throws IOException {
          Scanner sc = new Scanner(System.in);
          BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
          BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
  
          int t = Integer.parseInt(r.readLine());
          while(t-- > 0) {
              String s[] = r.readLine().split(" ");
              int a = Integer.parseInt(s[0]);
              int b = Integer.parseInt(s[1]);
  
              int result = (a-1)%4;
              if(result == 0)
                  result = (a-1);
              else if(result == 1)
                  result = 1;
              else if(result == 2)
                  result = a;
              else
                  result = 0;
  
              if(result == b)
                  System.out.println(a);
              else if((result^b) != a)    // include (result^b) as a term
                  System.out.println(a+1);
              else    // if(result^b) is a then x ^ (x^b^1) ^ 1   so took two more terms
                  System.out.println(a+2);
          }
  
      }
  }
  