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.lang.*;
  import java.io.*;
  import java.util.Arrays;
  
  public class Main
  {
      public static void main (String[] args) throws java.lang.Exception
      {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t;
          try {
              t = Integer.parseInt(br.readLine());
          } catch (Exception ignored){
              return;
          }
          while (t --> 0) {
              try {
                  int[] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                  if ((arr[1] & 1) == 1) {
                      if (arr[1] / 2 + 1 >= arr[0]) {
                          System.out.println(arr[1] / 2);
                      } else {
                          System.out.println(arr[1] % arr[0]);
                      }
                  } else {
                      if (arr[1] / 2 + 1 >= arr[0]) {
                          System.out.println(arr[1] /2 - 1);
                      } else {
                          System.out.println(arr[1] % arr[0]);
                      }
                  }
              } catch (Exception ignored){}
          }
      }
  } 