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;
  string dtb(long long a) {
    string s = "";
    for (long long i = 63; i >= 0; i--) {
      if (a & (1LL << i))
        s += '1';
      else
        s += '0';
    }
    return s;
  }
  void solve() {
    long long a, b;
    cin >> a >> b;
    long long len = a;
    long long ini, q;
    if (a % 4 == 0) {
      ini = 0;
      q = ini ^ b;
      if (ini == b)
        cout << len;
      else {
        if (q == a)
          cout << len + 2 << '\n';
        else
          cout << len + 1 << '\n';
      }
    } else if (a % 4 == 1) {
      ini = a - 1;
      q = ini ^ b;
      if (ini == b)
        cout << len;
      else {
        if (q == a)
          cout << len + 2;
        else
          cout << len + 1;
      }
    } else if (a % 4 == 2) {
      ini = 1;
      q = ini ^ b;
      if (ini == b)
        cout << len;
      else {
        if (q == a)
          cout << len + 2;
        else
          cout << len + 1;
      }
    } else {
      ini = a;
      q = ini ^ b;
      if (ini == b)
        cout << len;
      else {
        if (q == a)
          cout << len + 2;
        else
          cout << len + 1;
      }
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  