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;
  using ll = long long;
  using db = double;
  using pii = pair<int, int>;
  using vec = vector<int>;
  using veg = vector<pii>;
  template <typename T>
  T read() {
    T x = 0;
    bool f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void chkmax(T &x, const T &y) {
    x = x > y ? x : y;
  }
  template <typename T>
  void chkmin(T &x, const T &y) {
    x = x < y ? x : y;
  }
  const int N = 3e5 + 10;
  const int V = 2e5 + 10;
  int s[25][N];
  void solve() {
    int l = read<int>(), r = read<int>(), ans = 0;
    for (int i = (0); i <= (20); i++) {
      chkmax(ans, s[i][r] - s[i][l - 1]);
    }
    printf("%d\n", r - l + 1 - ans);
  }
  int main() {
    for (int i = (0); i <= (20); i++)
      for (int j = (1); j <= (V); j++) s[i][j] = s[i][j - 1] + (j >> i & 1);
    int T = read<int>();
    while (T--) solve();
    return 0;
  }
  