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;
  int main() {
    int t;
    int n, m;
    int i, j;
    int ans;
    vector<int> v1, v2;
    scanf("%d", &t);
    int flag = 0;
    while (t--) {
      v1.clear();
      v2.clear();
      flag = 0;
      scanf("%d%d", &n, &m);
      if (n == 0) {
        ans = m + 1;
        printf("%d\n", ans);
        continue;
      }
      if (n > m) {
        ans = 0;
        printf("%d\n", ans);
        continue;
      }
      int tmp = n;
      while (tmp > 0) {
        v1.push_back(tmp & 1);
        tmp = tmp >> 1;
      }
      tmp = m;
      while (tmp > 0) {
        v2.push_back(tmp & 1);
        tmp = tmp >> 1;
      }
      for (i = v1.size() - 1; i >= 0; i--) {
        if (v1[i] == v2[i]) continue;
        if (v1[i] > v2[i]) {
          flag = 1;
          break;
        }
      }
      int l;
      tmp = m + 1;
      j = 0;
      while (tmp > 0 && (tmp & 1) == 0) {
        tmp = tmp >> 1;
        j++;
      }
      ans = n ^ (m + 1);
      l = i + 1;
      ans = ans >> l;
      ans = ans << l;
      ans = ans >> j;
      ans = ans << j;
      printf("%d\n", ans);
    }
    return 0;
  }
  