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 int N = 1e5 + 5;
  int dfs(int l, int r, int B, int a, int b, int x) {
    if (l == r || a > b) return l;
    int mid = (1ll * l + r) >> 1;
    if (x >> B & 1) {
      if (b == r && a <= mid + 1)
        return dfs(l + (1 << B), r, B - 1, a + (1 << B), min(r, b + (1 << B)), x);
      else
        return dfs(l, l + (1 << B) - 1, B - 1, max(l, a - (1 << B)), b - (1 << B),
                   x);
    } else {
      if (l == a && b >= mid)
        return dfs(l + (1 << B), r, B - 1, max(a, l + (1 << B)), b, x);
      else
        return dfs(l, l + (1 << B) - 1, B - 1, a, min(l + (1 << B) - 1, b), x);
    }
  }
  void work() {
    int n, m;
    scanf("%d%d", &n, &m);
    printf("%d\n", dfs(0, (1 << 30) - 1, 29, 0, m, n));
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) work();
  }
  