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.io.*;
  
  public class HelloWorld{
  
      private static PrintWriter pw = new PrintWriter(System.out);
      private static InputReader sc = new InputReader();
      private static final int intmax = Integer.MAX_VALUE, intmin = Integer.MIN_VALUE;
   
      static class InputReader{
          private static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
          private static StringTokenizer tk;
   
          private void next()throws IOException{
              while(tk == null || !tk.hasMoreTokens())   
                  tk = new StringTokenizer(r.readLine());
          }
   
          private int nextInt()throws IOException{
              next();
              return Integer.parseInt(tk.nextToken());
          }
   
          private long nextLong()throws IOException{
              next();
              return Long.parseLong(tk.nextToken());
          }
   
          private String readString()throws IOException{
              next();
              return tk.nextToken();
          }
   
          private double nextDouble()throws IOException{
              next();
              return Double.parseDouble(tk.nextToken());
          }
   
          private int[] intArray(int n)throws IOException{
              next();
              int arr[] = new int[n];
   
              for(int i=0; i<n; i++)
                  arr[i] = nextInt();
   
              return arr;
          }
   
          private long[] longArray(int n)throws IOException{
              next();
              long arr[] = new long[n];
   
              for(int i=0; i<n; i++)
                  arr[i] = nextLong();
   
              return arr;
          }
      }
   
      public static void main(String args[])throws IOException{
          int t = sc.nextInt();
          // int t = 1;
          while(t-->0)    solve();
          pw.flush();
          pw.close();
      }
  
      public static void solve()throws IOException{
          long l = sc.nextLong(), r = sc.nextLong();
          
          long begin = Math.max(r/2, l);
          
          if(l > r/2)
              pw.println(r - l);
          else
              pw.println((r + 1)/2 - 1);
          
          /*if(r == begin * 2)
              pw.println(begin - 1);
          else
              pw.println(r - begin);*/
      }
  }