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 __gcd(long long a, long long b) {
    return b == 0 ? a : __gcd(b, a % b);
  }
  long long powmod(long long a, long long b) {
    long long res = 1;
    if (a >= 1000000007) a %= 1000000007;
    for (; b; b >>= 1) {
      if (b & 1) res = res * a;
      if (res >= 1000000007) res %= 1000000007;
      a = a * a;
      if (a >= 1000000007) a %= 1000000007;
    }
    return res;
  }
  static bool cmp(pair<long long, long long> &a, pair<long long, long long> &b) {
    return (a.second == b.second) ? a.first > b.first : a.second < b.second;
  }
  void printArray(vector<long long> arr) {
    for (long long i = 0; i < arr.size(); i++) {
      cout << arr[i] << " ";
    }
    cout << "\n";
  }
  bool prime[100000005];
  void SieveOfEratosthenes(long long n) {
    memset(prime, true, sizeof(prime));
    for (long long p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
  }
  long long lcm(long long a, long long b) { return (a / __gcd(a, b)) * b; }
  long long gcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long x1, y1;
    long long d = gcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return d;
  }
  bool find_any_solution(long long a, long long b, long long c, long long &x0,
                         long long &y0, long long &g) {
    g = gcd(abs(a), abs(b), x0, y0);
    if (c % g) {
      return false;
    }
    x0 *= c / g;
    y0 *= c / g;
    if (a < 0) x0 = -x0;
    if (b < 0) y0 = -y0;
    return true;
  }
  bool checkperfectsquare(long long n) {
    if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
      return true;
    } else {
      return false;
    }
  }
  bool check(long long a, long long b, long long c) {
    long long g = __gcd(a, b);
    if (c % g) {
      cout << "NO"
           << "\n";
      return false;
    }
    while (c % b && c > 0) c -= a;
    if (c >= 0)
      return true;
    else
      return false;
  }
  string test(string str, long long n) { return str.erase(n, 1); }
  void solve() {
    long long l, r, b;
    cin >> l >> r;
    if (l > (r / 2))
      cout << r % l << "\n";
    else
      cout << (r) % ((r / 2) + 1) << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
         << "\n";
    return 0;
  }
  