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;
  const long long int MOD = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int test = 1;
    cin >> test;
    while (test--) {
      long long int n, m, i, j;
      cin >> n >> m;
      long long int x = 0;
      if (m < n)
        cout << 0 << "\n";
      else {
        m++;
        long long int f = 0;
        for (i = 30; i > -1; i--) {
          long long int a, b;
          if (n & (1LL << i))
            a = 1;
          else
            a = 0;
          if (m & (1LL << i))
            b = 1;
          else
            b = 0;
          if (a == 0 && b == 0)
            ;
          else if (a == 1 && b == 0)
            x ^= (1LL << i), f = 1;
          else if (a == 1 && b == 1)
            x ^= (1LL << i);
          else {
            if (!f) x ^= (1LL << i);
          }
        }
        cout << (x ^ n) << "\n";
      }
    }
    cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
         << "\n";
    return 0;
  }
  