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 <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v);
  template <typename A, typename B>
  ostream &operator<<(ostream &cout, pair<A, B> const &p) {
    return cout << '(' << p.f << ',' << ' ' << p.s << ')';
  }
  template <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v) {
    cout << '[';
    for (int i = 0; i < v.size(); i++) {
      if (i) cout << ',';
      cout << v[i];
    }
    return cout << ']' << endl;
  }
  const long long mod = 1000000007;
  void solve() {
    long long a, b;
    cin >> a >> b;
    long long n = b / 2 + 1;
    cout << b % max(a, n) << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tc = 1;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  