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;
  template <class c>
  struct rge {
    c b, e;
  };
  template <class c>
  rge<c> range(c i, c j) {
    return rge<c>{i, j};
  }
  template <class c>
  auto dud(c* x) -> decltype(cerr << *x, 0);
  template <class c>
  char dud(...);
  struct debug {
    template <class c>
    debug& operator<<(const c&) {
      return *this;
    }
  };
  using ll = long long;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      ll l, r;
      cin >> l >> r;
      ll L = l, R = r, prospect = l;
      while (L <= R) {
        ll mid = (L + R) / 2;
        if (mid * 2 > r) {
          prospect = mid;
          R = mid - 1;
        } else
          L = mid + 1;
      }
      cout << r % prospect << '\n';
    }
  }
  