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 T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  void inp(T& x) {
    cin >> x;
  }
  template <class H, class... T>
  void inp(H& h, T&... t) {
    inp(h);
    inp(t...);
  }
  template <class A>
  void inp(vector<A>& x) {
    for (auto& a : x) inp(a);
  }
  void deb_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void deb_out(Head H, Tail... T) {
    cerr << ' ' << H;
    deb_out(T...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int mod = 1000000007;
  const char nl = '\n';
  void solve() {
    long long l, r;
    inp(l, r);
    if (r % 2 == 0) {
      if (l > r / 2 + 1)
        cout << r % l << nl;
      else
        cout << r % (r / 2 + 1) << nl;
    } else {
      if (l > (r + 1) / 2)
        cout << r % l << nl;
      else
        cout << r % ((r + 1) / 2) << nl;
    }
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int t = 1;
    inp(t);
    while (t--) solve();
    return 0;
  }
  