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 double PI = acos(-1.0);
  const long long MAX_N = 1e5 + 1;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long long LINF = 1e18;
  void solve() {
    long long l, r;
    cin >> l >> r;
    long long n{};
    long long out = (r - l + 1);
    r++;
    while ((1 << n) <= r) {
      long long a = (r >> (n + 1)) << n;
      if (r & (1 << n)) {
        a += r & ((1 << n) - 1);
      }
      long long b = (l >> (n + 1)) << n;
      if (l & (1 << n)) {
        b += l & ((1 << n) - 1);
      }
      out = min(out, (r - l) - (a - b));
      n++;
    }
    cout << out << '\n';
  }
  long long main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tc;
    cin >> tc;
    for (long long t = 1; t <= tc; t++) {
      solve();
    }
  }
  