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;
  long long int miracleAndSleeper(long long int l, long long int r) {
    long long int ans = 0;
    long long int s = l;
    long long int e = r;
    while (s != e) {
      long long int mid = (s + e) / 2;
      if (r % mid > ans) {
        e = mid;
        ans = r % mid;
      } else
        s = mid + 1;
    }
    return ans;
  }
  int main() {
    int test;
    cin >> test;
    while (test--) {
      long long int l, r;
      cin >> l >> r;
      cout << miracleAndSleeper(l, r) << endl;
    }
    return 0;
  }
  