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;
  int main() {
    int pp;
    cin >> pp;
    for (int jiaran = 1; jiaran <= pp; jiaran++) {
      int left, right, t = 1;
      cin >> left >> right;
      int p1 = 1, p2 = 2, temp = 0, p;
      int l, r, k1, k2, k;
      while (p2 <= right * 2) {
        p = 0;
        l = left / p1 + (left % p1 > 0);
        r = right / p1;
        if (r >= l) {
          p = (r - l);
          p /= 2;
          if (l % 2 > 0 && r % 2 == 0) p++;
          p *= p1;
          if (l % 2 == 0) p += l * p1 - left;
          if (r % 2 != 0) p += right - r * p1 + 1;
        } else {
          if (r % 2 > 0) p = right - left + 1;
        }
        if (p > temp) {
          temp = p;
        }
        p1 *= 2;
        p2 *= 2;
      }
      cout << right - left + 1 - temp << endl << endl;
    }
    return 0;
  }
  