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;
  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 f = 0;
    cerr << '{';
    for (auto& i : x) cerr << (f++ ? "," : ""), __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 adebug(int a[], int l, int r) {
    for (int i = l; i <= r; i++) {
      cerr << a[i] << " ";
    }
    cerr << '\n';
  }
  template <typename T, typename V>
  void mdebug(map<T, vector<V>> m) {
    for (auto x : m) {
      cerr << x.first << " : [ ";
      for (auto c : x.second) cerr << c << " ";
      cerr << "]" << '\n';
    }
  }
  template <class T>
  bool umin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool umax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class A>
  void read(vector<A>& v);
  template <class A, size_t second>
  void read(array<A, second>& a);
  template <class T>
  void read(T& x) {
    cin >> x;
  }
  void read(double& d) {
    string t;
    read(t);
    d = stod(t);
  }
  void read(long double& d) {
    string t;
    read(t);
    d = stold(t);
  }
  template <class H, class... T>
  void read(H& h, T&... t) {
    read(h);
    read(t...);
  }
  template <class A>
  void read(vector<A>& x) {
    for (auto& a : x) read(a);
  }
  template <class A, size_t second>
  void read(array<A, second>& x) {
    for (auto& a : x) read(a);
  }
  string to_string(char c) { return string(1, c); }
  string to_string(bool b) { return b ? "true" : "false"; }
  string to_string(const char* s) { return string(s); }
  string to_string(string s) { return s; }
  string to_string(vector<bool> v) {
    string res;
    for (int i = (0); ((1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size()));
         i += (1))
      res += char('0' + v[i]);
    return res;
  }
  template <size_t second>
  string to_string(bitset<second> b) {
    string res;
    for (int i = (0); ((1) > 0 ? i < (second) : i > (second)); i += (1))
      res += char('0' + b[i]);
    return res;
  }
  template <class T>
  string to_string(T v) {
    bool f = 1;
    string res;
    for (auto& x : v) {
      if (!f) res += ' ';
      f = 0;
      res += to_string(x);
    }
    return res;
  }
  template <class A>
  void write(A x) {
    cout << to_string(x);
  }
  template <class H, class... T>
  void write(const H& h, const T&... t) {
    write(h);
    write(t...);
  }
  void print() { write("\n"); }
  template <class H, class... T>
  void print(const H& h, const T&... t) {
    write(h);
    if (sizeof...(t)) write(' ');
    print(t...);
  }
  struct PH {
    size_t operator()(const pair<int, int>& x) const {
      size_t ans = 0;
      for (int i = 0; i < x.first; i++) ans += x.second;
      return ans;
    }
  };
  template <class T>
  void offset(long long o, T& x) {
    x += o;
  }
  template <class T>
  void offset(long long o, vector<T>& x) {
    for (auto& a : x) offset(o, a);
  }
  template <class T, size_t second>
  void offset(long long o, array<T, second>& x) {
    for (auto& a : x) offset(o, a);
  }
  template <class T>
  void fk(T a) {
    print(a);
    exit(0);
  }
  long const M = 1e9 + 7;
  const long long INF = 1e18;
  const long long mxN = 1e6, di[4] = {1, 0, -1, 0}, dj[4] = {0, -1, 0, 1};
  long long get(long long n, long long k) {
    long long res = (n >> (k + 1LL)) << k;
    if (((n >> k) & 1LL) != 0) res += n & ((1LL << k) - 1LL);
    return res;
  }
  void solve() {
    long long l, r;
    read(l, r);
    long long ans = 1e9;
    for (long long i = (0); ((1) > 0 ? i < (20) : i > (20)); i += (1)) {
      long long d = get(r + 1, i) - get(l, i);
      umin(ans, r - l + 1 - d);
    }
    print(ans);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long T = 1;
    read(T);
    for (long long _ = (0); ((1) > 0 ? _ < (T) : _ > (T)); _ += (1)) {
      solve();
    }
    return 0;
  }
  