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.util.*;
  import java.io.*;
  
  public class Solution {
  
      public static void main(String[] args) throws IOException {
  
          Scanner in = new Scanner(System.in);
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringBuffer out = new StringBuffer();
  
          int T = in.nextInt();
  
          int max = 300_000;
          int xor[] = new int[max+1];
          for(int i=1; i<=max; i++) {
              xor[i]^=(xor[i-1]^i);
          }
  
          OUTER:
          while (T-- > 0) {
  
              int a = in.nextInt(),
                  b = in.nextInt();
  
              int ans = -1;
              if(xor[a-1]==b) {
                  ans = a;
              } else if((xor[a-1]^b)!=a) {
                  ans = a+1;
              } else {
                  ans = a+2;
              }
              out.append(ans+"\n");
          }
  
          System.out.print(out);
      }
  
      private static void sort(List<Integer> list) {
          list.stream().sorted();
      }
  
      private static long gcd(long a, long b) {
          if (a==0)
              return b;
          return gcd(b%a, a);
      }
  
      private static int toInt(String s) {
          return Integer.parseInt(s);
      }
  
      private static long toLong(String s) {
          return Long.parseLong(s);
      }
  }