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>
  using namespace std;
  long long n, m;
  vector<long long> holes;
  long long _2[64];
  long long masked(long long mask) {
    long long result = 0;
    for (int i = 0; i < 30; i++) {
      if (mask & _2[i]) {
        result = result + (_2[holes[i]]);
      }
    }
    return result;
  }
  int main() {
    _2[0] = 1;
    for (int i = 1; i <= 62; i++) {
      _2[i] = _2[i - 1] * 2;
    }
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      if (n > m) {
        cout << "0\n";
        continue;
      }
      holes.clear();
      for (int i = 0; i < 61; i++) {
        if (_2[i] & n)
          continue;
        else
          holes.push_back(i);
        if (holes.size() == 30) break;
      }
      long long l = 0;
      long long r = _2[30] + 5;
      long long ans = -1;
      while (l <= r) {
        long long mid = (l + r) / 2;
        if (masked(mid) + n > m) {
          ans = masked(mid);
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  