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 long long iinf = 1e9 + 10;
  const long long inf = 1ll << 60;
  const long long mod = 1e9 + 7;
  void GG() {
    cout << "0\n";
    exit(0);
  }
  long long mpow(long long a, long long n, long long mo = mod) {
    long long re = 1;
    while (n > 0) {
      if (n & 1) re = re * a % mo;
      a = a * a % mo;
      n >>= 1;
    }
    return re;
  }
  long long inv(long long b, long long mo = mod) {
    if (b == 1) return b;
    return (mo - mo / b) * inv(mo % b, mo) % mo;
  }
  const long long maxn = 2e5 + 5;
  long long cnt[20][maxn];
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ;
    for (long long i = 0; i < 20; ++i) {
      for (long long j = 1; j <= maxn - 1; ++j) {
        cnt[i][j] = cnt[i][j - 1] + ((j & (1 << i)) != 0);
      }
    }
    long long t;
    cin >> t;
    while (t--) {
      long long l, r;
      cin >> l >> r;
      long long re = 0;
      for (long long i = 0; i < 20; ++i) {
        re = max(re, (__typeof__(re))(cnt[i][r] - cnt[i][l - 1]));
      }
      cout << (r - l + 1) - re << '\n';
    }
  }
  