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.StringTokenizer;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  public class Main 
  {
          public static void main(String... args) throws IOException
          {
                  int tt;
                  //Scanner sc= new Scanner(System.in);
            BufferedReader inf=new BufferedReader(new InputStreamReader(System.in));      
            StringTokenizer st=new StringTokenizer(inf.readLine());      
                  tt=Integer.parseInt(st.nextToken());
                  int l,r;
                   while((tt--)>0)
                   { 
                  st=new StringTokenizer(inf.readLine());
                   l=Integer.parseInt(st.nextToken());
                    r=Integer.parseInt(st.nextToken());
                  //l=Integer.parseInt(inf.readLine());
                  //r=Integer.parseInt(inf.readLine());
                  System.out.println((((r-1)/2 +1)>=l)?((r-1)/2):(r%l));  
                   }
          }
  }