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 mod = 1e9 + 7;
  long long power1(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    if (b % 2 == 0)
      return (power1((a * a), b / 2));
    else
      return (power1((a * a), b / 2) * a);
  }
  long long power(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    if (b % 2 == 0)
      return (power((a * a) % mod, b / 2)) % mod;
    else
      return (power((a * a) % mod, b / 2) * a) % mod;
  }
  long long fact(long long n) {
    if (n == 0) return 1;
    if (n == 1) return 1;
    return (fact(n - 1) * n) % mod;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  vector<bool> prime(1e7, true);
  void sieve(long long n) {
    for (int p = 2; p * p <= n; p++) {
      if (prime[p]) {
        for (int i = p * 2; i <= n; i += p) prime[i] = false;
      }
    }
  }
  void swap(long long &a, long long &b) {
    long long t = b;
    b = a;
    a = t;
  }
  long long invmod(long long n) { return power(n, mod - 2); }
  int main() {
    long long t;
    cin >> t;
    for (; t > 0; t--) {
      long long l, r;
      cin >> l >> r;
      long long ans = INT_MAX;
      if (l == r) {
        cout << 0 << "\n";
        continue;
      }
      for (long long div = 1; div <= r; div *= 2) {
        long long base = div * 2;
        long long sl = l / base;
        long long sr = r / base;
        long long cnt = 0;
        if (sl == sr) {
          cnt = min(base / 2, r - sr * base + 1) - min(base / 2, l - sl * base);
          ans = min(ans, cnt);
          continue;
        }
        if (sr > sl) {
          cnt += (sr - sl - 1) * (base / 2);
        }
        cnt += (base / 2 - (min(base / 2, l - sl * base)));
        cnt += (min(base / 2, r - sr * base + 1));
        ans = min(ans, cnt);
      }
      cout << ans << "\n";
    }
    return 0;
  }
  