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 <class T, class V>
  T power(T t, V v) {
    T second = 1;
    while (v--) second *= t;
    return second;
  }
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << " ";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    int n = v.size();
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      n--;
      cerr << (n ? ", " : " ");
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    int n = v.size();
    cerr << "{ ";
    for (auto p : v) {
      cerr << "[";
      _print(p.first);
      cerr << "]=";
      _print(p.second);
      n--;
      cerr << (n ? ", " : " ");
    }
    cerr << "}";
  }
  vector<long long> p(25);
  void go() {
    long long l, r;
    cin >> l >> r;
    long long ans = 0;
    for (int i = 0; i < 24; i++) {
      long long temp = 0;
      long long a = 0;
      if (l >= p[i]) {
        a = l / p[i];
        if (a % 2 == 1) {
          long long z = l - a * p[i];
          a = (a / 2) * p[i] + z;
        } else {
          a = (a / 2) * p[i];
        }
      }
      long long b = 0;
      if (r >= p[i]) {
        b = r / p[i];
        if (b % 2 == 1) {
          long long z = r - b * p[i] + 1;
          b = (b / 2) * p[i] + z;
        } else {
          b = (b / 2) * p[i];
        }
      }
      ans = max(ans, b - a);
    }
    cout << (r - l + 1) - ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    p[0] = 1;
    for (int i = 1; i < 25; i++) {
      p[i] = p[i - 1] * 2;
    }
    cerr << "p"
         << ": ";
    _print(p);
    cerr << "\n";
    ;
    while (t--) {
      go();
    }
    return 0;
  }
  