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;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << '\n'; }
  template <typename Up, typename... Down>
  void dbg_out(Up U, Down... D) {
    cerr << ' ' << U;
    dbg_out(D...);
  }
  const int N = 2e5 + 10;
  int pref[N][30];
  bool check_bit(int x, int pos) { return (x & (1 << pos)); }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    for (int i = 1; i < N; i++) {
      for (int j = 0; j < 30; j++) {
        if (check_bit(i, j)) {
          pref[i][j] = (pref[i - 1][j] + 1);
        } else
          pref[i][j] = pref[i - 1][j];
      }
    }
    int tt;
    cin >> tt;
    while (tt--) {
      int l, r;
      cin >> l >> r;
      int arr[30] = {0};
      for (int i = 0; i < 30; i++) {
        arr[i] = pref[r][i] - pref[l - 1][i];
      }
      int tot = r - l + 1;
      int ans = INT_MAX;
      for (int i = 0; i < 30; i++) {
        ans = min(tot - arr[i], ans);
      }
      cout << ans << '\n';
    }
    return 0;
  }
  