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: #include <bits/stdc++.h>
  using namespace std;
  int powmod(int x, int y, int p) {
    int res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y % 2 == 1) res = (res * x) % p;
      y /= 2;
      x = (x * x) % p;
    }
    return res;
  }
  string dectobin(int x) {
    string s = "";
    while (x > 0) {
      int t = x % 2;
      s.push_back(t + '0');
      x /= 2;
    }
    reverse(s.begin(), s.end());
    if (s.compare("") == 0)
      return "0";
    else
      return s;
  }
  int long long bintodec(string s) {
    int long long ans = 0;
    int long long n = s.size();
    for (int long long i = n - 1; i >= 0; i--) {
      if (s[i] == '1') ans += pow(2, n - i - 1);
    }
    return ans;
  }
  int find(int k, int n) { return ((n & (1 << (k - 1))) >> (k - 1)); }
  int long long arr[3 * 100000 + 1];
  int32_t main() {
    arr[0] = 0;
    for (int long long i = 0; i < (3 * 100000 + 1); i++) arr[i] = i ^ arr[i - 1];
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int long long a, b;
      cin >> a >> b;
      int long long k = 0;
      k = arr[a - 1];
      int long long sum = a;
      string s = "";
      for (int long long i = 0; i < (40); i++) {
        int long long u = 0, v = 0;
        u = b & (1LL << i);
        v = k & (1LL << i);
        if (u == v)
          s = "0" + s;
        else
          s = "1" + s;
      }
      int long long l = bintodec(s);
      if (l == a)
        sum += 2;
      else if (l == 0)
        sum += 0;
      else
        sum += 1;
      cout << sum << "\n";
    }
    return 0;
  }
  