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>
  long long mod = 1000000007;
  using namespace std;
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % mod;
      b >>= 1;
      a = (a * a) % mod;
    }
    return res;
  }
  bool isprime(long long n) {
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) return 0;
    }
    return 1;
  }
  long long pref[200005][31];
  void precalculate() {
    long long N = 2e5 + 5;
    for (long long i = 1; i < N; i++) {
      for (int j = 0; j <= 30; j++) {
        if (i & (1 << j)) pref[i][j] = 1;
        pref[i][j] += pref[i - 1][j];
      }
    }
  }
  void solve() {
    long long l, r;
    cin >> l >> r;
    long long total = r - l + 1;
    long long ans = 1e10;
    for (int j = 0; j <= 30; j++) {
      long long curr = (pref[r][j] - pref[l - 1][j]);
      ans = min(ans, total - curr);
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    precalculate();
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  