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;
  vector<char> prime();
  vector<int> faster_prime();
  long long int power();
  long long int powermod();
  void solution() {
    long long int l, r;
    cin >> l >> r;
    long long int ans = r / 2 - 1;
    if (r % 2) {
      ans = r / 2;
    }
    if (l > r / 2) {
      ans = r % l;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
      solution();
    }
    return 0;
  }
  vector<char> prime(int n) {
    vector<char> c(n + 1, '1');
    c.at(0) = '0';
    c.at(1) = '0';
    for (int i = 2; i * i <= n; i++) {
      if (c[i] == '1') {
        for (int j = i * i; j <= n; j += i) {
          c.at(j) = '0';
        }
      }
    }
    return c;
  }
  vector<int> faster_prime(int n) {
    vector<int> factor(n + 1, 0);
    vector<int> prime;
    for (int i = 2; i < n; i++) {
      if (factor[i] == 0) {
        factor.at(i) = i;
        prime.push_back(i);
      }
      for (int j = 0; j < prime.size() && prime[j] <= i && i * prime[j] <= n;
           j++) {
        factor.at(i * prime[j]) = prime[j];
      }
    }
    return prime;
  }
  long long int power(long long int a, long long int n) {
    long long int pow = 1;
    for (long long int i = n; i > 0; i >>= 1) {
      if (i & 1) {
        pow *= a;
      }
      a *= a;
    }
    return pow;
  }
  long long int powermod(long long int a, long long int n, long long int m) {
    long long int pow = 1;
    a %= m;
    for (long long int i = n; i > 0; i >>= 1) {
      if (i & 1) {
        pow = pow * a % m;
      }
      a = a * a % m;
    }
    return pow;
  }
  