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 edit(vector<int> &skill, int n) {
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      skill.push_back(x);
    }
  }
  void editl(vector<long long> &skill, long long n) {
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      skill.push_back(x);
    }
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  bool isGoogles = 0;
  bool testcase = 1;
  void achilles() {
    long long a, b;
    cin >> a >> b;
    if (b % 2) {
      if (a <= b / 2 + 1) {
        cout << b / 2 << "\n";
        return;
      } else {
        cout << b % a << "\n";
        return;
      }
    } else {
      if (a <= b / 2 + 1) {
        cout << b / 2 - 1 << "\n";
        return;
      } else {
        cout << b % a << "\n";
        return;
      }
    }
  }
  int main() {
    ios ::sync_with_stdio(0);
    cin.tie(0);
    long long t = 1, w = 1;
    if (testcase) cin >> t;
    while (t--) {
      if (isGoogles) {
        cout << "Case #" << w << ": ";
      }
      achilles();
      w++;
    }
    return 0;
  }
  