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;
  int main() {
    int t;
    cin >> t;
    vector<vector<int>> v(200001, vector<int>(19, 0));
    for (int i = 1; i < 200001; ++i) {
      long long int p = 1;
      int c = 0;
      for (; p < 6 * 1e5; p *= 2) {
        if (i & p) {
          v[i][c] = v[i - 1][c] + 1;
        } else {
          v[i][c] = v[i - 1][c];
        }
        c++;
      }
    }
    while (t--) {
      int l, r;
      cin >> l >> r;
      int temp = INT_MIN;
      for (int i = 0; i < 19; ++i) {
        temp = max(temp, v[r][i] - v[l - 1][i]);
      }
      cout << r - l + 1 - temp << endl;
    }
    return 0;
  }
  