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.*;
  
  public class B {
    static final int MAXN = (int)3e5 + 5;
    static int[] f = new int[MAXN];
    
    public static void main(String[] args) {
      int t;
      Scanner sc = new Scanner(System.in);
      t = sc.nextInt();
      
      f[0] = 0;
      for (int i = 1; i < MAXN; ++i) {
        f[i] = f[i-1] ^ i;
      }
      
      while(t > 0) {
        t--;
        int n = sc.nextInt();
        int m = sc.nextInt();
  
        if (n > 0 && f[n-1] == m) {
          System.out.println(n);
        } else if ((f[n-1] ^ m) == n) {
          System.out.println(n+2);
        } else {
          System.out.println(n+1);
        }
  
      }
    }
  }