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>
  template <class T>
  bool chmax(T &a, const T &b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  using namespace std;
  string to_string(const string &s) { return '"' + s + '"'; }
  string to_string(const char *s) { return to_string((string)s); }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A>
  string to_string(A v) {
    bool first = 1;
    string res = "{";
    for (const auto &x : v) {
      if (!first) res += ", ";
      first = 0;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  using ll = long long;
  using pii = pair<int, int>;
  using vi = vector<int>;
  using db = double;
  using ldb = long double;
  const int maxn = 300000;
  const int mod = 998244353;
  const int inf = 0x3f3f3f3f;
  int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
      auto cal = [&](int x, int d) {
        x++;
        int N = 1 << (d + 1);
        int c = x / N;
        int res = x % N >= N / 2 ? N / 2 : x % N;
        res += c * (N / 2) - 1;
        return res;
      };
      int l, r;
      scanf("%d%d", &l, &r);
      int ans = inf;
      for (auto i = 0; i <= 20; i++) chmin(ans, cal(r, i) - cal(l - 1, i));
      printf("%d\n", ans);
    }
    return 0;
  }
  