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;
  void err(istream_iterator<string> it) {}
  template <typename T, typename... Args>
  void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << " ";
    err(++it, args...);
  }
  const long long MX = 2e5 + 105;
  bool checkbit(int mask, int bit) { return (mask & (1 << bit)); }
  int setbit(int mask, int bit) { return (mask | (1 << bit)); }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, m;
      cin >> n >> m;
      int ans = 2e9, now = 0;
      for (int bit = 29; bit >= 0; bit--) {
        if (!checkbit(m, bit)) {
          int cur = now;
          if (!checkbit(n, bit)) {
            cur = setbit(cur, bit);
            ans = min(ans, cur);
          } else {
            ans = min(ans, cur);
            now = setbit(now, bit);
          }
        } else {
          if (!checkbit(n, bit)) {
            now = setbit(now, bit);
          }
        }
      }
      cout << ans << endl;
    }
  }
  