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>
  const long long MOD = 1e9 + 7;
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T &a, const T &b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int first = 0;
    cerr << '{';
    for (auto &i : x) cerr << (first++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  void test_case() {
    int a, b;
    cin >> a >> b;
    int cnt = 0;
    long long ans = a;
    for (long long i = 1; i <= b; i *= 2) {
      ans = i;
      if (ans >= a) break;
    }
    for (int i = a; i <= b; i++) {
      if ((i & ans) == 0)
        cnt++;
      else
        ans &= i;
    }
    ans /= 2;
    int cnt2 = 0;
    for (int i = a; i <= b; i++) {
      if ((ans & i) == 0)
        cnt2++;
      else
        ans &= i;
    }
    ckmin(cnt, cnt2);
    int mn = (b - a + 1) / 2;
    if (a % 2 == 0 && b % 2 == 0) mn++;
    cout << min(mn, cnt) << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int te = 1;
    cin >> te;
    while (te--) {
      test_case();
    }
  }
  