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 T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &i : v) cin >> i;
    return in;
  }
  template <typename T>
  ostream &operator<<(ostream &out, vector<T> &v) {
    for (auto &i : v) cout << i << " ";
    return out;
  }
  void solve() {
    long long int i, j, n, a, b, c, d, m, k, sum = 0, ans = 0, cnt = 0, r = 0,
                                             e = 0, x = 0, y = 0;
    string str, s1, om, s;
    map<long long int, long long int> mp, mp1;
    set<long long int> st;
    cin >> a >> b;
    c = b - a;
    b--;
    if ((b + 1) >= 2 * a)
      ans = b / 2;
    else
      ans = c;
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long int tc;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  