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: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class a {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          int tc = Integer.parseInt(br.readLine());
  
          for (int tt = 0; tt < tc; tt++) {
              st = new StringTokenizer(br.readLine());
              int l = Integer.parseInt(st.nextToken());
              int r = Integer.parseInt(st.nextToken());
  
              int answer = 0;
              int leftMost = r/2-1;
              int rightMost = r/2+1;
  
              for (int div = leftMost; div <= rightMost; div++) {
                  if (div > r || div < l) continue;
                  answer = Math.max(answer, r % div);
              }
              answer = Math.max(answer, r%l);
  
              System.out.println(answer);
  
          }
      }
  
  
      private static int gcd(int a, int b) {
          if (a % b == 0) return b;
          else return gcd(b, a % b);
      }
  }