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 int;
  using ull = unsigned long long int;
  using ld = long double;
  constexpr ll MAX = 2000000000000000000;
  constexpr ld PI = 3.14159265358979;
  constexpr ll MOD = 0;
  ld dotorad(ld K) { return PI * K / 180.0; }
  ld radtodo(ld K) { return K * 180.0 / PI; }
  mt19937 mt;
  void randinit() {
    srand((unsigned)time(NULL));
    mt = mt19937(rand());
  }
  ll c(ll N, ll i) {
    N++;
    ll p = (1LL << i), q = (1LL << (i + 1));
    ll ans = N / q * p;
    N %= q;
    ans += min(N, p);
    return ans;
  }
  int main() {
    ll T;
    cin >> T;
    for (ll t = 0; t < T; t++) {
      ll L, R;
      cin >> L >> R;
      ll ans = MAX;
      for (ll i = 0; i < 20; i++) {
        ans = min(ans, c(R, i) - c(L - 1, i));
      }
      cout << ans << endl;
    }
  }
  