description: You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  \operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.
  
  The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).
  
  Output
  
  For each test case, print a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  3 5
  4 6
  3 2
  69 696
  123456 654321
  
  
  Output
  
  
  4
  3
  0
  640
  530866
  
  Note
  
  In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.
  
  In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.
  
  In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
solution: #include <bits/stdc++.h>
  template <class C>
  C& mini(C& a, C b) {
    if (b < a) a = b;
    return a;
  }
  template <class C>
  C& maxi(C& a, C b) {
    if (a < b) a = b;
    return a;
  }
  template <typename T>
  void read(std::vector<T>& A) {
    for (T& x : A) std::cin >> x;
  }
  template <typename T>
  void readv(std::vector<T>& A) {
    int N;
    std::cin >> N;
    A.resize(N);
    read(A);
  }
  using namespace std;
  int getmask(int N) {
    int b = 1;
    while (b <= N) b <<= 1;
    return b - 1;
  }
  int solve(int N, int M) {
    if (N == 0) return M + 1;
    if (N > M) return 0;
    int m = getmask(N);
    int left = M & ~m, right = M & m;
    if (left == 0) {
      assert(right);
      if (((right + 1) & right) == 0) return M + 1;
      int b = (m + 1) >> 1;
      assert(N & b);
      assert(M & b);
      return solve(N ^ b, M ^ b);
    } else {
      if ((right & m) == m) return M + 1;
      return left ^ solve(N, M ^ left);
    }
  }
  int main(int argc, const char* argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    cout.setf(ios::fixed);
    int T;
    cin >> T;
    for (int t = 1; t <= T; t++) {
      int N, M;
      cin >> N >> M;
      cout << solve(N, M) << '\n';
    }
    return 0;
  }
  