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 int N = 1e6 + 1;
  const int rN = 1e5 + 1;
  constexpr long double PI = 3.141592653589793238462;
  template <typename T1, typename T2>
  istream &operator>>(istream &istream, pair<T1, T2> &p) {
    return (istream >> p.first >> p.second);
  }
  template <typename T>
  istream &operator>>(istream &istream, vector<T> &v) {
    for (auto &it : v) cin >> it;
    return istream;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
    return (ostream << p.first << " " << p.second);
  }
  template <typename T>
  ostream &operator<<(ostream &ostream, const vector<T> &c) {
    for (auto &it : c) cout << it << " ";
    return ostream;
  }
  bool prime[N];
  void process() {
    for (int i = 2; i < N; i++) prime[i] = true;
    prime[0] = prime[1] = false;
    for (int i = 2; i * i <= N; i++) {
      if (prime[i] == true) {
        for (int j = i * i; j < N; j += i) {
          prime[j] = false;
        }
      }
    }
  }
  long long power(long long a, long long n, long long num) {
    long long res = a;
    long long x = 1;
    while (n) {
      if (n % 2 == 1) x = (x * res) % num;
      res = (res * res) % num;
      n /= 2;
    }
    return x % num;
  }
  long long inverse(long long n, long long num) { return power(n, num - 2, num); }
  int gcd(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
  int dp[31][200001];
  void solve() {
    int l, r;
    cin >> l >> r;
    int ans = INT_MAX;
    for (int i = 0; i < 31; i++) {
      ans = min(ans, dp[i][r] - dp[i][l - 1]);
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    memset((dp), 0, sizeof((dp)));
    int temp;
    int idx;
    for (int i = 1; i < 200001; i++) {
      for (int j = 0; j < 31; j++) {
        dp[j][i] += dp[j][i - 1] + (!(i & (1 << j)));
      }
    }
    int tt;
    cin >> tt;
    while (tt--) solve();
    return 0;
  }
  