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;
  void solved_by_satyapsr13() {
    long long int n, k, ans = 0, l, count = 0, sum = 0, mn = INT_MAX,
                        mx = INT_MIN;
    long long int r;
    cin >> l >> r;
    long long int test = r / 2;
    if (test >= l) {
      r & 1 ? cout << test : cout << test - 1;
      cout << "\n";
      return;
    } else {
      cout << r % l;
      cout << "\n";
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long int Test_Cases;
    cin >> Test_Cases;
    while (Test_Cases--) {
      solved_by_satyapsr13();
    }
    return 0;
  }
  