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.util.Scanner;
  public class Main 
  {
          public static void main(String... args)
          {
                  int tt;
                  Scanner sc= new Scanner(System.in);
                  tt=sc.nextInt();
  
                  int i,j,l,r;
                   while((tt--)>0)
                   { 
                  l=sc.nextInt();r=sc.nextInt();
                  /*int mini=Integer.MIN_VALUE;
                  for(i=l;i<=r;i++)
                  {
                          mini=Math.max(mini,r%i);
                  }*/
                  System.out.println((((r-1)/2 +1)>=l)?((r-1)/2):(r%l));  
                   }
          }
  }
  