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;
  const long long inf = 1e15;
  long long l, r;
  void Input() { cin >> l >> r; }
  void Solve() {
    if (r <= 10) {
      long long Res = 0;
      for (long long a = l; a <= r; ++a) {
        for (long long b = l; b <= a; ++b) Res = max(Res, a % b);
      }
      cout << Res;
    } else {
      long long m = max(r / 2 + 1, l);
      cout << max(r % l, r % m);
    }
    cout << '\n';
  }
  int main() {
    if (fopen("trash.inp", "r"))
      freopen("trash.inp", "r", stdin), freopen("trash.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int Test;
    cin >> Test;
    while (Test--) {
      Input();
      Solve();
    }
  }
  