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 PV(vector<long long> v) {
    for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
    cout << "\n";
  }
  void PVV(vector<pair<long long, long long> > v) {
    for (long long i = 0; i < (long long)v.size(); i++)
      cout << v[i].first << " " << v[i].second << "\n";
  }
  void PA(long long v[], long long n, long long x = 0) {
    for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
    cout << "\n";
  }
  void IN(long long a[], long long n, long long x = 0) {
    for (long long i = x; i < n + x; i++) cin >> a[i];
  }
  inline void op() {}
  const long long M = 1e9 + 7;
  const long long IM = 1e18 + 37;
  const long long N = 2e5 + 10;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    op();
    long long t;
    cin >> t;
    while (t--) {
      long long l, r;
      cin >> l >> r;
      if (l == r) {
        cout << (0) << "\n";
        continue;
      }
      long long a = (r + 1) / 2;
      long long ans = (a + a - 1) % a;
      if (a < l) {
        ans = (r % l);
      }
      cout << (ans) << "\n";
    }
    return 0;
  }
  