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;
  using namespace std;
  const double eps = 1e-6;
  const int int_inf = 2 * 1e9;
  const long long ll_inf = 8 * 1e18;
  const long double PI = acosl(-1.0);
  template <typename T>
  using pair2 = pair<T, T>;
  using pii = pair<int, int>;
  using pli = pair<long long, int>;
  using pll = pair<long long, long long>;
  istream& operator>>(istream& in, pii& i) {
    in >> i.first >> i.second;
    return in;
  }
  template <class T>
  istream& operator>>(istream& in, vector<T>& a) {
    for (auto& i : a) {
      in >> i;
    }
    return in;
  }
  template <class T>
  ostream& operator<<(ostream& out, vector<T>& a) {
    for (auto i : a) {
      out << i << " ";
    }
    out << '\n';
    return out;
  }
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long myRand(long long B) { return (unsigned long long)rng() % B; }
  long long powmod(long long a, long long b, long long p) {
    a %= p;
    if (a == 0) return 0;
    long long product = 1;
    while (b > 0) {
      if (b & 1) {
        product *= a;
        product %= p;
        --b;
      }
      a *= a;
      a %= p;
      b /= 2;
    }
    return product;
  }
  long long factor[35];
  long long inv(long long a, long long p) { return powmod(a, p - 2, p); }
  long long nCk(long long n, long long k, long long p) {
    return ((factor[n] * inv(factor[k], p) % p) * inv(factor[n - k], p)) % p;
  }
  const int SIZE = 2 * 1e5 + 8;
  vector<vector<int>> prefs;
  void solve() {
    int r, l;
    cin >> l >> r;
    vector<int> curr = prefs[r];
    vector<int> curl = prefs[l - 1];
    int ans = 1e9;
    for (int i = 0; i < 32; ++i) {
      ans = min(ans, curr[i] - curl[i]);
    }
    cout << ans << '\n';
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    prefs.resize(SIZE, vector<int>(33, 0));
    for (int i = 1; i < 2 * 1e5 + 3; ++i) {
      prefs[i] = prefs[i - 1];
      for (int j = 0; j <= 31; ++j) {
        if ((i & (1 << j)) == 0) {
          prefs[i][j]++;
        }
      }
    }
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  