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>
  inline void read(T &a) {
    T w = 1;
    a = 0;
    char ch = getchar();
    for (; ch < '0' || ch > '9'; ch = getchar())
      if (ch == '-') w = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
      a = (a << 3) + (a << 1) + (ch ^ 48);
    a *= w;
  }
  template <typename T>
  inline void ckmax(T &a, T b) {
    a = a > b ? a : b;
  }
  template <typename T>
  inline void ckmin(T &a, T b) {
    a = a < b ? a : b;
  }
  int main() {
    int cases;
    read(cases);
    while (cases--) {
      int n, m;
      read(n);
      read(m);
      m++;
      int Ans = 0, fl = 0;
      for (int i = 30; i >= 0; --i) {
        if ((n & (1 << i)) && !(m & (1 << i))) fl = 1;
        if (!fl && !(n & (1 << i)) && (m & (1 << i))) Ans += 1 << i;
      }
      cout << Ans << endl;
    }
  }
  