description: You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?
  
  A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.
  
  Output
  
  For each test case, output a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  1 2
  2 8
  4 5
  1 5
  100000 200000
  
  
  Output
  
  
  1
  3
  0
  2
  31072
  
  Note
  
  In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.
  
  In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
solution: #include <bits/stdc++.h>
  using namespace std;
  const double pi = 3.14159265358979323846;
  const int M = 1e6 + 1;
  vector<int> adj[M];
  vector<long long> ans;
  bool gone[M];
  int pre1[300000][32];
  void pre() {
    for (long long i = 1; i <= 200000; i++) {
      long long num1 = i;
      long long pos1 = 0;
      while (num1 != 0) {
        pre1[i][pos1] = num1 % 2 + pre1[i - 1][pos1];
        pos1++;
        num1 = num1 / 2;
      }
    }
  }
  void solve() {
    long long num1, num2;
    cin >> num1 >> num2;
    int max1 = INT_MIN;
    for (long long i = 0; i < 32; i++) {
      max1 = max(pre1[num2][i] - pre1[num1 - 1][i], max1);
    }
    cout << num2 - num1 + 1 - max1 << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    pre();
    long long freq = 1;
    cin >> freq;
    while (freq--) {
      solve();
    }
  }
  