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;
  void fast_io() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  template <typename A, typename B>
  ostream &operator<<(ostream &cout, pair<A, B> const &p) {
    return cout << "(" << p.first << ", " << p.second << ")";
  }
  template <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v) {
    cout << "[";
    for (int i = 0; i < (int)v.size(); i++) {
      if (i) cout << ", ";
      cout << v[i];
    }
    return cout << "]";
  }
  mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
  int main() {
    fast_io();
    int test;
    cin >> test;
    vector<vector<int> > pref(3e5, vector<int>(20));
    for (int i = 1; i <= 3e5; i++) {
      pref[i] = pref[i - 1];
      for (int j = 0; j < 20; j++) pref[i][j] += (i >> j) % 2;
    }
    for (int tc = 0; tc < test; tc++) {
      int l, r;
      cin >> l >> r;
      vector<int> bits = pref[r];
      for (int j = 0; j < 20; j++) bits[j] -= pref[l - 1][j];
      int best = 0;
      for (int j = 0; j < 20; j++) best = max(best, bits[j]);
      cout << r - l + 1 - best << "\n";
    }
  }
  