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 mi(long long n, long long m) {
    long long pw = n % m;
    long long ex = m - 2;
    long long ans = 1;
    while (ex) {
      if (ex & 1) ans = ans * pw % m;
      pw = pw * pw % m;
      ex >>= 1;
    }
    return ans % 1000000007;
  }
  long long pw(long long a, long long n) {
    if (n < 0) return 0;
    if (n == 0) return 1;
    long long pw = a;
    long long ex = n;
    long long ans = 1;
    while (ex) {
      if (ex & 1) ans = ans * pw;
      pw = pw * pw;
      ex >>= 1;
    }
    return ans;
  }
  long long pwm(long long a, long long n) {
    long long pw = a % 1000000007;
    long long ex = n;
    long long ans = 1;
    while (ex) {
      if (ex & 1) ans = (ans % 1000000007 * pw % 1000000007) % 1000000007;
      pw = (pw % 1000000007 * pw % 1000000007) % 1000000007;
      ex >>= 1;
    }
    return ans % 1000000007;
  }
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long max2(long long a, long long b) { return a > b ? a : b; }
  long long min2(long long a, long long b) { return a < b ? a : b; }
  long long max3(long long a, long long b, long long c) {
    return max2(a, max2(b, c));
  }
  long long min3(long long a, long long b, long long c) {
    return min2(a, min2(b, c));
  }
  long long ceil(long long a, long long b) { return a / b + (a % b != 0); }
  void _print(long long t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(double t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long dx[] = {1, 0, -1, 0};
  long long dy[] = {0, 1, 0, -1};
  long long X[] = {2, 1, -1, -2, -2, -1, 1, 2};
  long long Y[] = {1, 2, 2, 1, -1, -2, -2, -1};
  struct comp {
    bool operator()(const pair<long long, long long>& a,
                    const pair<long long, long long>& b) const {
      if (a.second > b.second) return true;
      return false;
    }
  };
  static bool comp(string a, string b) {
    string aa = a + b;
    string bb = b + a;
    return aa.compare(bb) < 0;
  }
  void solve() {
    long long l, r;
    cin >> l >> r;
    long long x = (r + 1) / 2;
    if (r % 2 == 0) x++;
    if (x >= l) {
      cout << r % x;
    } else {
      cout << r % l;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
      cout << "\n";
    }
  }
  