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.awt.dnd.DragSource;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.lang.reflect.Array;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.List;
  import java.util.Map;
  import java.util.Set;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  public class A {
  
      static int one(int n){
          int ans=0;
          while(n>0){
              ans++;
              n = n&(n-1);
          }
          return ans;
      }
      public static void main(String[] args) throws IOException {
          FastScanner s = new FastScanner();
          int n = (int)3e5+7;
          int[] ar = new int[n];
          for (int i = 1; i <n ; i++) {
              ar[i] = ar[i - 1] ^ (i - 1);
          }
          int t = s.nextInt();
  //        int t=1;
          while(t-->0){
              int a = s.nextInt();
              int b = s.nextInt();
              int v = ar[a];
              if(v==b) System.out.println(a);
              else if((v^b)==a) System.out.println(a+2);
              else System.out.println(a+1);
          }
      }
  
  
      static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
      static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
  
      static void sortr(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l,Collections.reverseOrder());
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sort(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }