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 <typename AA>
  inline void CMax(AA &u, AA v) {
    if (v > u) u = v;
  }
  template <typename AA>
  inline void CMin(AA &u, AA v) {
    if (v < u) u = v;
  }
  const double eps = 1E-8;
  const int Inf = 0x3f3f3f3f, Mod = 1E9 + 7;
  const long long Lnf = 0x3f3f3f3f3f3f3f3f;
  template <typename AA>
  void read(AA &x) {
    int s = 0;
    char c;
    for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
      ;
    c == '-' ? (s = 1, x = 0) : (x = c - '0');
    while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
    if (s) x = -x;
  }
  template <typename AA, typename... Args>
  void read(AA &t, Args &...args) {
    read(t);
    read(args...);
  }
  template <typename AA>
  void print(AA k) {
    int num = 0;
    static char stk[23];
    if (k < 0) {
      putchar('-');
      k = -k;
    }
    while (k) stk[++num] = k % 10 + 48, k /= 10;
    if (!num)
      putchar('0');
    else
      while (num) putchar(stk[num--]);
    putchar('\n');
  }
  inline int Log2(int x) {
    if (!x)
      return 0;
    else
      return 31 - __builtin_clz(x);
  }
  void gch(char &ch) {
    do {
      ch = getchar();
    } while (ch == ' ' || ch == '\n');
  }
  long long Pow(long long d, long long m) {
    long long f = 1;
    while (m) {
      if (m & 1) f = f * d % Mod;
      d = d * d % Mod;
      m >>= 1;
    }
    return f;
  }
  int T = 1, N, M, S, Q, K;
  int s[300030];
  void solve() {
    int a, b;
    read(a, b);
    if (s[a - 1] == b)
      print(a);
    else if ((s[a - 1] ^ b) != a)
      print(a + 1);
    else
      print(a + 2);
  }
  int main() {
    for (int a = (1); a <= (300000); ++a) s[a] = s[a - 1] ^ a;
    read(T);
    while (T--) {
      solve();
    }
    return 0;
  }
  