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;
  const long long mod = 1000000007;
  struct mint {
    long long x;
    mint() : x(0) {}
    mint(long long x) : x((x % mod + mod) % mod) {}
    mint& fix() {
      x = (x % mod + mod) % mod;
      return *this;
    }
    mint operator-() const { return mint(0) - *this; }
    mint& operator+=(const mint& a) {
      if ((x += a.x) >= mod) x -= mod;
      return *this;
    }
    mint& operator-=(const mint& a) {
      if ((x += mod - a.x) >= mod) x -= mod;
      return *this;
    }
    mint& operator*=(const mint& a) {
      (x *= a.x) %= mod;
      return *this;
    }
    mint operator+(const mint& a) const { return mint(*this) += a; }
    mint operator-(const mint& a) const { return mint(*this) -= a; }
    mint operator*(const mint& a) const { return mint(*this) *= a; }
    bool operator<(const mint& a) const { return x < a.x; }
    bool operator==(const mint& a) const { return x == a.x; }
  };
  template <typename T>
  void show(vector<T> v) {
    for (auto it : v) {
      cout << it << " ";
    }
    cout << "\n";
  }
  void solve() {
    int r, l;
    std::cin >> l >> r;
    if (l == r) {
      std::cout << 0 << "\n";
      return;
    }
    int left = (r + 1) / 2;
    if (left < l) left = l;
    int remain = r % left;
    int rr = (r - 1) % left;
    std::cout << std::max(remain, rr) << "\n";
  }
  int main() {
    int t = 1;
    std::cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  