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 max2(long long x, long long y) {
    if (x > y)
      return x;
    else
      return y;
  }
  long long min2(long long x, long long y) {
    if (x < y)
      return x;
    else
      return y;
  }
  unsigned long long min3(unsigned long long x, unsigned long long y) {
    if (x < y)
      return x;
    else
      return y;
  }
  long double min(long double x, long double y) {
    if (x < y)
      return x;
    else
      return y;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long parent(long long i) { return (i - 1) / 2; }
  long long left(long long i) { return i * 2 + 1; }
  long long right(long long i) { return 2 * i + 2; }
  bool within(long long x, long long l, long long r) {
    return ((x >= l) && (x <= r)) || ((x <= l) && (x >= r));
  }
  void solve(long long t) {
    unsigned long long l, r;
    cin >> l >> r;
    long long ans = r / 2;
    if (ans >= l) {
      while (2 * (ans + 1) - 1 <= r) ans++;
      cout << ans - 1 << endl;
    } else
      cout << r % l << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    int q = 0;
    cin >> t;
    while (t--) {
      q++;
      solve(q);
    }
    return 0;
  }
  