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;
  template <class T>
  using maxheap = priority_queue<T, vector<T>, less<T>>;
  template <class T>
  using minheap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  void setmax(T& a, T b) {
    a = max(a, b);
  };
  template <typename T>
  void setmin(T& a, T b) {
    a = min(a, b);
  };
  template <typename T>
  bool in(T lo, T v, T hi) {
    return lo <= v && v <= hi;
  };
  template <typename T>
  ostream& operator<<(ostream& out, vector<T>& v) {
    out << "[";
    int n = v.size();
    for (int i = 0; i < n; i++) out << v[i] << (i + 1 < n ? ", " : "");
    out << "]";
    return out;
  }
  struct timestamp {
    string algo;
    long double A, B;
    timestamp(string _algo) {
      algo = _algo;
      A = B = clock();
    }
    void print() {
      long double B_ = clock();
      long double section = (B_ - B) / CLOCKS_PER_SEC;
      long double cumulative = (B_ - A) / CLOCKS_PER_SEC;
      cout << algo << " profile(seconds): section: " << section
           << " cumulative: " << cumulative << '\n'
           << flush;
      B = B_;
    }
  };
  template <long long MOD>
  struct ModInt {
    long long x = 0;
    ModInt() { x = 0; }
    ModInt(long long _x) {
      x = _x;
      if (x >= MOD || x <= -MOD) x %= MOD;
      if (x < 0) x += MOD;
    }
    friend ModInt operator+(ModInt a, ModInt b) { return ModInt(a.x + b.x); }
    friend ModInt operator-(ModInt a, ModInt b) { return ModInt(a.x - b.x); }
    friend ModInt operator*(ModInt a, ModInt b) { return ModInt(a.x * b.x); }
    friend ModInt operator/(ModInt a, ModInt b) { return a * inv(b); }
    friend ModInt operator+=(ModInt& a, ModInt b) { return a = a + b; }
    friend ModInt power(ModInt a, long long b) {
      if (b == 0) return 1;
      ModInt c = power(a, b / 2);
      return c * c * ((b & 1) ? a : 1);
    }
    friend ModInt inv(ModInt a) { return power(a, MOD - 2); }
    friend ostream& operator<<(ostream& os, const ModInt& a) {
      os << a.x;
      return os;
    }
    friend istream& operator>>(istream& is, ModInt& a) {
      long long y;
      is >> y;
      a = ModInt(y);
      return is;
    }
  };
  using mint = ModInt<998244353>;
  using vm = vector<mint>;
  using vvm = vector<vm>;
  const int N = 3e5 + 10;
  long long X[N];
  void precalc() {
    for (int i = 1; i < N; i++) X[i] = X[i - 1] ^ i;
  }
  struct Solution {
    void solve() {
      long long m, x;
      cin >> m >> x;
      long long y = X[m - 1];
      if (y == x) {
        cout << m << '\n';
      } else if ((y ^ x) == m) {
        cout << (m + 2) << '\n';
      } else {
        cout << (m + 1) << '\n';
      }
    }
  };
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    precalc();
    int t = 1;
    cin >> t;
    while (t--) {
      Solution second;
      second.solve();
    }
    return 0;
  }
  