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;
  template <typename... Args>
  void logger(string vars, Args &&...values) {
    cout << vars << " = ";
    string delim = "";
    (..., (cout << delim << values, delim = ", "));
  }
  template <typename T>
  void print(vector<T> &collection, bool newline = true) {
    for (auto &item : collection) {
      cout << item << " ";
    }
    if (newline) cout << "\n";
  }
  template <typename T>
  void print(T _data, bool newline = true) {
    cout << _data;
    if (newline) cout << "\n";
  }
  long long f;
  long long n, w, k, x, m, t, p;
  long long cnt = 0;
  long long getSetBitsFromOneToN(long long r, long long l) {
    long long two = 2, ans = INT_MAX;
    long long n1 = r, n2 = l;
    l--;
    long long n = r;
    while (n) {
      long long curr = 0;
      curr += (r / two) * (two >> 1);
      if ((r & (two - 1)) > (two >> 1) - 1)
        curr += (r & (two - 1)) - (two >> 1) + 1;
      n >>= 1;
      if (true) {
        curr -= (l / two) * (two >> 1);
        if ((l & (two - 1)) > (two >> 1) - 1)
          curr -= (l & (two - 1)) - (two >> 1) + 1;
      }
      two <<= 1;
      ans = min(ans, n1 - n2 + 1 - curr);
    }
    return ans;
  }
  long long helper(long long l, long long r) {
    long long ans = INT_MAX;
    for (long long bit = 31; bit >= 0; bit--) {
      long long curr = 0;
      for (long long num = l; num <= r; num++) {
        if (num & (1 << bit)) curr++;
      }
      ans = min(ans, r - l + 1 - curr);
    }
    return ans;
  }
  void Solve() {
    long long l, r;
    cin >> l >> r;
    cout << getSetBitsFromOneToN(r, l);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long test;
    cin >> test;
    for (long long _ = 0; _ < test; _++) {
      Solve();
      cout << "\n";
    }
  }
  