description: You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.
  
  Input
  
  Each test contains multiple test cases.
  
  The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.
  
  The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).
  
  Output
  
  For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.
  
  Example
  
  Input
  
  
  4
  1 1
  999999999 1000000000
  8 26
  1 999999999
  
  
  Output
  
  
  0
  1
  12
  499999999
  
  Note
  
  In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.
  
  In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
solution: #include <bits/stdc++.h>
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    std::cerr << name << " : " << arg1 << "\n";
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  std::pair<int, int> operator+(std::pair<int, int> a, std::pair<int, int> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::pair<long long, long long> operator+(std::pair<long long, long long> a,
                                            std::pair<long long, long long> b) {
    return {a.first + b.first, a.second + b.second};
  }
  std::ostream& operator<<(std::ostream& out, std::pair<int, int> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::ostream& operator<<(std::ostream& out, std::pair<long long, long long> a) {
    out << a.first << " " << a.second << "\n";
    return out;
  }
  std::istream& operator>>(std::istream& in, std::pair<int, int>& a) {
    in >> a.first >> a.second;
    return in;
  }
  std::istream& operator>>(std::istream& in, std::pair<long long, long long>& a) {
    in >> a.first >> a.second;
    return in;
  }
  long long bpow(long long x, long long v) {
    if (v == 0) return 1 % 1000000007;
    if (v % 2 == 1) return (x * bpow(x, v - 1)) % 1000000007;
    long long r = bpow(x, v / 2);
    return (r * r) % 1000000007;
  }
  using namespace std;
  void solve() {
    long long l, r;
    cin >> l >> r;
    long long b = max(l, r / 2 + r % 2);
    cout << max(r % b, max(l, (r - 1)) % b) << "\n";
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  