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>
  using namespace std;
  struct _ {
    ios_base::Init i;
    _() {
      cin.sync_with_stdio(0);
      cin.tie(0);
      cout << fixed << setprecision(10);
    }
  } ___;
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    cerr << name << " : " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << "  ";
    __f(comma + 1, args...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long int my_rand(long long int l, long long int r) {
    return uniform_int_distribution<long long int>(l, r)(rng);
  }
  signed main() {
    long long int __, _ = 0;
    cin >> __;
    for (; _++ < __;) {
      long long int l, r;
      cin >> l >> r;
      long long int lo = l, hi = r;
      while (lo < hi) {
        long long int mid = (lo + hi + 1) >> 1;
        if (mid * 2 - 1 >= l and mid * 2 - 1 <= r)
          lo = mid;
        else
          hi = mid - 1;
      }
      if (lo * 2 - 1 >= l and lo * 2 - 1 <= r) {
        cout << lo - 1 << '\n';
      } else
        cout << r % l << '\n';
    }
  }
  