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 MAX = 5005, INF = 1e15;
  long long MOD = 998244353;
  template <typename T>
  void c_min(T &a, const T b) {
    if (b <= a) a = b;
  }
  template <typename T>
  void c_max(T &a, const T b) {
    if (a <= b) a = b;
  }
  template <typename T>
  T mexp(T a, long long e) {
    T res = 1;
    while (e) {
      if (e & 1) res *= a;
      a *= a;
      e >>= 1;
    }
    return res;
  }
  void test_case(long long tc = 0) {
    int l, r;
    cin >> l >> r;
    int x = r / 2 + 1;
    int ans = r % l;
    if (x >= l) {
      ans = max(ans, r % x);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long T = 1, tc = 0;
    cin >> T;
    for (; tc < T; tc++) {
      test_case(tc);
    }
    return 0;
  }
  