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>
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  #pragma GCC optimize("Ofast")
  using namespace std;
  long long int mod = 1000000007LL;
  template <class A = long long int>
  void Out(vector<A> a) {
    for (long long int i = (long long int)0; i < (long long int)a.size(); i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
  template <class A = long long int>
  void In(vector<A> &a, long long int N) {
    A cur;
    for (long long int i = (long long int)0; i < (long long int)N; i++) {
      cin >> cur;
      a.push_back(cur);
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int T;
    cin >> T;
    for (long long int t = (long long int)0; t < (long long int)T; t++) {
      long long int N, M;
      cin >> N >> M;
      if (N > M) {
        cout << 0 << endl;
        continue;
      }
      M++;
      long long int D = (long long int)log2(4LL * M);
      long long int y = 0LL;
      for (long long int i = D - 1; i >= 0; i--) {
        long long int dN = N & (1LL << i);
        long long int dM = M & (1LL << i);
        if (dN != 0LL && dM == 0LL) {
          break;
        }
        if (dN == 0LL && dM != 0LL) {
          y += (1LL << i);
        }
      }
      cout << y << endl;
    }
    return 0;
  }
  