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;
  const long long mod = 998244353;
  const long long N = 3e5 + 5;
  const int base = 350;
  const int base2 = 311;
  long long n, m, t, k, T, tong, a[N], b[N], c[N], d[N], ans, dp[N], lab[N], h[N],
      P[N][20];
  vector<long long> adj[N], kq;
  long long pw(long long k, long long n) {
    long long total = 1;
    for (; n; n >>= 1) {
      if (n & 1) total = total * k % mod;
      k = k * k % mod;
    }
    return total;
  }
  string s;
  void dfs(long long u, long long p) {
    for (int i = 1; i < 20; i++) P[u][i] = P[P[u][i - 1]][i - 1];
    for (long long v : adj[u]) {
      if (v == p) continue;
      h[v] = h[u] + 1;
      P[v][0] = u;
      dfs(v, u);
    }
  }
  long long lca(long long u, long long v) {
    if (h[u] < h[v]) swap(u, v);
    long long log = log2(h[u]) + 1;
    for (int i = log; i >= 0; i--) {
      if (h[u] >= h[v] + (1 << i)) u = P[u][i];
    }
    if (u == v) return u;
    for (int i = log; i >= 0; i--) {
      if (P[u][i] && P[u][i] != P[v][i]) {
        u = P[u][i];
        v = P[v][i];
      }
    }
    return P[u][0];
  }
  long long findp(long long u) { return lab[u] < 0 ? u : lab[u] = findp(lab[u]); }
  void hop(long long u, long long v) {
    lab[u] += lab[v];
    lab[v] = u;
  }
  void sol() {
    n = 3e5;
    for (int i = 1; i <= n; i++) a[i] = (a[i - 1] ^ i);
    cin >> t;
    while (t-- > 0) {
      cin >> n >> m;
      if ((m ^ a[n - 1]) == 0)
        cout << n << '\n';
      else if ((m ^ a[n - 1]) != n)
        cout << n + 1 << '\n';
      else
        cout << n + 2 << '\n';
    }
  }
  int main() {
    if (fopen("INTERNET"
              ".in",
              "r")) {
      freopen(
          "INTERNET"
          ".in",
          "r", stdin);
      freopen(
          "INTERNET"
          ".out",
          "w", stdout);
    }
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int ntest = 1;
    while (ntest-- > 0) sol();
  }
  