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 int N = 2e5 + 5;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      long long l, r;
      cin >> l >> r;
      map<long long, long long> m;
      for (long long i = 0; i <= log2(r); i++) {
        long long x = i + 1;
        long long y = 1 << x;
        long long ans = 0;
        ans += ((r + 1) / y) * (y / 2);
        ans -= ((l) / y) * (y / 2);
        long long a = l % y;
        long long b = (r + 1) % y;
        if (b) {
          if (b <= (y / 2))
            ans += b;
          else
            ans += (y / 2);
        }
        if (a) {
          if (a <= (y / 2))
            ans -= a;
          else
            ans -= (y / 2);
        }
        m[i] += ans;
      }
      long long res = INT_MAX;
      for (auto i : m) {
        if (i.second < res) res = i.second;
      }
      cout << res << endl;
    }
    return 0;
  }
  