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;
  long long int msbPos(long long n) {
    long long int msb_p = -1;
    while (n) {
      n = n >> 1;
      msb_p++;
    }
    return msb_p;
  }
  long long binpow(long long a, long long b) {
    if (b == 0) return 1;
    long long res = binpow(a, b / 2);
    if (b % 2)
      return res * res * a;
    else
      return res * res;
  }
  void solve() {
    long long l, r;
    cin >> l >> r;
    long long a = msbPos(l);
    long long b = msbPos(r);
    if (a == b) {
      cout << 0 << endl;
      return;
    } else {
      vector<long long> s(b - a + 1);
      s[0] = (binpow(2, a)) - (l % (binpow(2, a)));
      s[b - a] = (r % (binpow(2, b))) + 1;
      for (long long i = 1; i < b - a; i++) {
        s[i] = binpow(2, a + i);
      }
      vector<long long> v(b + 1, 0);
      for (long long i = 0; i < b - a + 1; i++) {
        for (long long j = 0; j < a + i; j++) {
          long long d = binpow(2, j + 1);
          v[j] += (s[i] / d) * (d / 2);
          if (s[i] % d > d / 2) {
            v[j] += s[i] % d - d / 2;
          }
        }
        v[a + i] += s[i];
      }
      sort(v.rbegin(), v.rend());
      cout << r - l + 1 - v[0] << endl;
    }
  }
  int main() {
    long long int t;
    cin >> t;
    while (t > 0) {
      solve();
      t--;
    }
  }
  