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 mod = 1e9 + 7;
  const long long N = 4e6 + 6;
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long powmod(long long a, long long b) {
    long long ans = 1;
    while (b > 0) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  long long readn() {
    long long sum = 0, f = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      sum = sum * 10 + (ch - '0');
      ch = getchar();
    }
    return sum * f;
  }
  long long t, l, r, p;
  int main() {
    t = readn();
    while (t--) {
      l = readn();
      r = readn();
      p = r - l;
      if (p >= l)
        cout << r / 2 - (r % 2 == 0);
      else
        cout << p;
      cout << endl;
    }
  }
  