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.BufferedReader; 
  import java.io.IOException; 
  import java.io.InputStreamReader; 
  import java.util.Scanner; 
  import java.util.StringTokenizer;
  import java.util.*;
  public class codeforcesA{
     static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
   
          public FastReader() 
          { 
              br = new BufferedReader(new
                       InputStreamReader(System.in)); 
          } 
   
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException  e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
   
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
   
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
   
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
   
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
          } 
     } 
     public static void main(String args[]){
       FastReader sc=new FastReader();
       StringBuilder sb=new StringBuilder();
       int t=sc.nextInt();
       while(t-->0){
         int b=sc.nextInt();
         int a=sc.nextInt();
         int ans=0;
         b=Math.max(((a/2)+1),b);
         System.out.println(a%b);
       }
    } 
  }