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;
  int Power(int base, int exp) {
    int ans = 1;
    while (exp > 0) {
      if (exp & 1) ans = (1LL * ans * base) % 1000000007;
      exp = exp >> 1;
      base = (1LL * base * base) % 1000000007;
    }
    return ans;
  }
  void solve() {
    long long int l, r;
    cin >> l >> r;
    long long int x = r / 2;
    if (x >= l) {
      if (r % 2 == 0)
        cout << x - 1 << '\n';
      else
        cout << x << '\n';
    } else
      cout << r % l << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    int q = 1;
    cin >> q;
    while (q--) solve();
  }
  