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;
  bool debug_cerr = false;
  vector<long long> mf, mi;
  long long POW(long long x, long long y);
  long long MI(long long x, long long mod);
  void facts(long long x, long long mod);
  long long C(long long x, long long y, long long mod);
  long long bpow(long long x, long long y, long long mod);
  long long POW(long long x, long long y) { return bpow(x, y, (long long)1e18); }
  long long MI(long long x, long long mod = 1000000007) {
    return bpow(x, mod - 2, mod);
  }
  void facts(long long x, long long mod = 1000000007) {
    mf = mi = vector<long long>(x + 1);
    mf[0] = 1;
    for (long long i = 1; i <= x; i++) {
      mf[i] = (i * mf[i - 1]) % mod;
    }
    mi[x] = MI(mf[x], mod);
    for (long long i = x - 1; i >= 0; i--) {
      mi[i] = ((i + 1) * mi[i + 1]) % mod;
    }
    return;
  }
  long long C(long long x, long long y, long long mod = 1000000007) {
    if (x < y) return 0;
    long long n1 = mf[x];
    long long n2 = mi[y];
    long long n3 = mi[x - y];
    long long res = (n1 * n2) % mod;
    res = (res * n3) % mod;
    return res;
  }
  long long bpow(long long x, long long y, long long mod = 1000000007) {
    x %= mod;
    long long res = 1;
    while (y > 0) {
      if (y & 1) {
        res = res * x % mod;
      }
      x = x * x % mod;
      y >>= 1;
    }
    return res;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long q;
    cin >> q;
    for (long long _ = 1; _ <= q; _++) {
      long long a, b;
      cin >> a >> b;
      long long ans = a;
      long long d = a % 4;
      long long x = 0;
      a--;
      while (d--) {
        x = (x ^ (a));
        a--;
      }
      if (x == b) {
        cout << ans << '\n';
      } else {
        if ((x ^ b) == ans) {
          cout << ans + 2 << '\n';
        } else {
          cout << ans + 1 << '\n';
        }
      }
    }
  }
  