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 int N = 3e5 + 15;
  const int mod = 1e9 + 7;
  inline int read() {
    int ret = 0, op = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') op = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      ret = ret * 10 + ch - '0';
      ch = getchar();
    }
    return ret * op;
  }
  inline long long readll() {
    long long ret = 0, op = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') op = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      ret = ret * 10 + ch - '0';
      ch = getchar();
    }
    return ret * op;
  }
  void write(int x) {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline int Max(const int e, const int f) { return e > f ? e : f; }
  int xr[N];
  int main() {
    for (register int i = 1; i <= 3e5; i++) xr[i] = xr[i - 1] ^ i;
    int t = read();
    while (t--) {
      int a = read();
      int b = read();
      if (xr[a - 1] == b)
        printf("%d\n", a);
      else {
        if ((xr[a - 1] ^ b) == a) {
          printf("%d\n", a + 2);
        } else
          printf("%d\n", a + 1);
      }
    }
    return 0;
  }
  