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 t_max = 1e9 + 7;
  const long long MAXX = long long(1e18);
  const int INF = int(-1e9);
  const double PI = 3.1415926535;
  void read() {}
  template <typename T1>
  void read(T1 v[], int n) {
    for (int i = 0; i < n; i++) cin >> v[i];
  }
  template <typename T2>
  void read(vector<T2>& v, int n) {
    for (int i = 0; i < v.size(); i++) cin >> v[i];
  }
  template <typename First, typename... Rest>
  void read(First& first, Rest&... rest) {
    cin >> first;
    read(rest...);
  }
  template <typename T3>
  void print(vector<T3>& v, int n) {
    for (int i = 0; i < v.size(); i++) cout << v[i] << '\n';
  }
  template <typename T4>
  void print(T4 v[], int n) {
    for (int i = 0; i < n; i++) cout << v[i] << '\n';
  }
  long long binpow(long long a, long long n, long long k) {
    long long res = 1;
    while (n)
      if (n & 1) {
        res *= a;
        res %= k;
        n--;
      } else {
        a *= a;
        a %= k;
        n >>= 1;
      }
    return res;
  }
  const int M = 200001, L = 23;
  int arr[M][L];
  void solve() {
    int l, r;
    cin >> l >> r;
    int maxx = 0;
    for (int i = 0; i < L; i++) maxx = max(maxx, arr[r][i] - arr[l - 1][i]);
    cout << r - l + 1 - maxx << '\n';
  }
  int main() {
    for (int i = 0; i < L; i++) arr[0][i] = 0;
    for (int i = 1; i < M; i++) {
      for (int j = 0; j < L; j++) arr[i][j] = arr[i - 1][j];
      int k = i;
      int pos = 0;
      while (k != 0) {
        arr[i][pos] += k % 2;
        k /= 2;
        pos++;
      }
    }
    cin.tie(0);
    cout.tie(0);
    std::ios_base::sync_with_stdio(0);
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; i++) {
      solve();
    }
  }
  