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.lang.*;
  import java.io.*;
  
  //author: ankit_bit
  
  public class Main {
  
  
      static FastReader fr = new FastReader();
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) throws java.lang.Exception {
  
          int testcases = fr.nextInt();
          long arr[]= new long[300001];
          arr[0]=0;
          for(int i=1;i<300001;i++){
              arr[i]=arr[i-1]^i;
          }
          while (testcases-- > 0) {
  
              solve(arr);
          }
          System.out.println(sb);
  
      }
  
      public static void solve(long arr[]) {
        int mex= fr.nextInt();
        int xor= fr.nextInt();
        long cnt=0, currxor=0;
  //      for(int i=0;i<mex;i++){
  //          cnt++;
  //          currxor^=i;
  //      }
        cnt=mex;
        currxor=arr[(int) (cnt-1)];
        if(currxor==xor){
            sb.append(cnt);
        }
        else{
            if((currxor^xor)==mex)
            sb.append(cnt+2);
            else{
                sb.append(cnt+1);
            }
        }
        sb.append("\n");
      }
  
      static boolean isPrime(int x) {
          for (int d = 2; d * d <= x; d++) {
              if (x % d == 0)
                  return false;
          }
          return true;
      }
  
      static int lowerBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  high = mid - 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static int upperBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  low = mid + 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static long mult(long a, long b, long mod) {
          long res = (a * b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long add(long a, long b, long mod) {
          long res = (a + b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long gcd(long a, long b) {
          if (b == 0) return a;
          return gcd(b, a % b);
      }
  
      static long lcm(long a, long b) {
          return (a / gcd(a, b)) * b;
      }
  
      static long fact[] = new long[100001], M = 1000000007;
  
      static long ncr(int n, int r) {
          if (r > n) return 0;
          return fact[n] * power(fact[r], M - 2) % M * power(fact[n - r], M - 2) % M;
      }
  
      static long power(long a, long b) {
          long res = 1;
          while (b != 0) {
              if (b % 2 == 1) res = res * a % M;
              b >>= 1;
              a = a * a % M;
          }
          return res;
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i = 0; i < a.length; i++) {
              l.add(a[i]);
          }
          Collections.sort(l);
          for (int i = 0; i < l.size(); i++)
              a[i] = l.get(i);
      }
  
  }
  
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
  
      public FastReader() {
          br = new BufferedReader(new
                  InputStreamReader(System.in));
      }
  
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int nextInt() {
          return Integer.parseInt(next());
      }
  
      long nextLong() {
          return Long.parseLong(next());
      }
  
      double nextDouble() {
          return Double.parseDouble(next());
      }
  
      String nextLine() {
          String str = "";
          try {
              str = br.readLine();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
  }
  