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.*;
  import java.lang.*;
  // import java.math.*;
  import java.io.*;
  public class Main
  {
      public static int mod=(int)Math.pow(10,9) + 7;
      public static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
      public static PrintWriter ot=new PrintWriter(System.out);
      public static int[] take_arr(int n){
          int a[]=new int[n];
          try {
              String s[]=br.readLine().trim().split(" ");
              for(int i=0;i<n;i++)
              a[i]=Integer.parseInt(s[i]);
          } catch (Exception e) {
              e.printStackTrace();
          }
          return a;
      }
  	public static void main (String[] args) throws java.lang.Exception
  	{
  		try{
              int t=Integer.parseInt(br.readLine().trim());
              while(t-->0)
              {
                  String qq[]=br.readLine().trim().split(" ");
                  int l=Integer.parseInt(qq[0]);
                  int r=Integer.parseInt(qq[1]);
                  if(l<=r/2)
                  ot.println(r%(r/2 + 1));
                  else
                  ot.println(r%l);
              }
              ot.close();
              br.close();
              
          }catch(Exception e){
              e.printStackTrace();
              return;
          }
      }
      
  }
      