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 com.company;
  
  import java.math.*;
  import java.util.*;
  import java.io.*;
  
  public class Main {
  
      public static void main(String[] args) {
          FastScanner in = new FastScanner();
  
          int t = in.nextInt();
          while (t-- > 0) {
              int a = in.nextInt();
              int b = in.nextInt();
              int x;
              a--;
              if (a % 4 == 0) {
                  x = a;
              } else if (a % 4 == 1) {
                  x = 1;
              } else if (a % 4 == 2) {
                  x = a + 1;
              } else {
                  x = 0;
              }
              if (x != b) {
                  if ((x ^ b) == a + 1) {
                      System.out.println(a + 3);
                  } else {
                      System.out.println(a + 2);
                  }
              } else {
                  System.out.println(a + 1);
              }
          }
      }
  
      static class FastScanner {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public FastScanner() {
              reader = new BufferedReader(new InputStreamReader(System.in), 32768);
              tokenizer = null;
          }
  
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public double nextDouble() {
              return Double.parseDouble(next());
          }
  
          public String nextLine() {
              try {
                  return reader.readLine();
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
          }
      }
  }
  