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;
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  long long n = 200001;
  class Solution {
   public:
    vector<vector<int>> arr;
    void precalculate() {
      this->arr = vector<vector<int>>(n, vector<int>(20, 0));
      for (int i = 1; i < n; ++i) {
        int num = i;
        int k = 0;
        while (num > 0) {
          arr[i][k] = arr[i - 1][k] + (num & 1);
          num = num >> 1;
          ++k;
        }
      }
    }
  
   public:
    int solve(int start, int end) {
      int max_offsets = 0;
      for (int i = 0; i < 20; ++i) {
        max_offsets = max(max_offsets, arr[end][i] - arr[start - 1][i]);
      }
      return (end - start + 1 - max_offsets);
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    Solution s = Solution();
    s.precalculate();
    while (t--) {
      int l, r;
      cin >> l >> r;
      cout << s.solve(l, r) << endl;
    }
    return 0;
  }
  