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;
  template <typename T>
  using MinHeap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using MaxHeap = priority_queue<T>;
  const long double PI = 3.14159265358979323846;
  const long double E = 2.718281828459;
  long long MOD = (1ll << 31) - 1;
  long long M(long long n) { return (n % MOD + MOD) % MOD; }
  long long pow2(long long n) { return 1ll << n; }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  int main(int argn, char **argv) {
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
      long long n, m;
      cin >> n >> m;
      long long ans = 0, cannot = 0;
      for (long long(bit) = (31); (bit) >= (0); (bit)--) {
        long long bt = 1ll << bit;
        long long lim = (ans ^ n) | (bt - 1);
        if (m >= lim) {
          ans += bt;
        }
      }
      if ((n ^ ans) <= m) ans++;
      ;
      cout << ans << endl;
    }
    return 0;
  }
  