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.PrintWriter;
  import java.math.BigInteger;
  import java.util.*;
  import java.util.stream.Collectors;
  
  import static java.lang.Math.max;
  import static java.lang.Math.min;
  
  public class Main {
  
      public static void main(String[] args) {
          try(PrintWriter pw = new PrintWriter(System.out)) {
              try (Scanner sc = new Scanner(System.in)) {
                  solve(pw, sc);
              }
          }
      }
  
      private static void solve(PrintWriter out, Scanner in) {
          preCalc();
          int t = in.nextInt();
          in.nextLine();
          for (int i = 0; i < t; i++) {
              solveNext(in, out);
          }
      }
  
      private static void preCalc() {
  
      }
  
      private static void solveNext(Scanner in, PrintWriter out) {
          int l = in.nextInt();
          int r = in.nextInt();
          out.println(r % max(l, r / 2 + 1));
      }
  
  }
  