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.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.*;
  import java.io.*;
  public class Main {
      public static void main(String args[])throws Exception{
          Input sc=new Input();
          StringBuilder sb=new StringBuilder();
          int t=sc.readInt();
          for(int f=0;f<t;f++){
              int d[]=sc.readArray();
              int l=d[0];
              int r=d[1];
              int n=r/2;
              if(l>n){
                 sb.append(r%l);
              }else{
                  sb.append(r%(n+1));
              }
              sb.append("\n");
          }
          System.out.print(sb);
      }
  }
  class Input{
      BufferedReader br;
      StringTokenizer st;
      Input(){
          br=new BufferedReader(new InputStreamReader(System.in));
          st=new StringTokenizer("");
      }
      public int[] readArray() throws Exception{
          st=new StringTokenizer(br.readLine());
          int a[]=new int[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Integer.parseInt(st.nextToken());
          }
          return a;
      }
      public long[] readArrayLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          long a[]=new long[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Long.parseLong(st.nextToken());
          }
          return a;
      }
      public int readInt() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Integer.parseInt(st.nextToken());
      }
      public long readLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Long.parseLong(st.nextToken());
      }
      public String readString() throws Exception{
          return br.readLine();
      }
      public int[][] read2dArray(int n,int m)throws Exception{
          int a[][]=new int[n][m];
          for(int i=0;i<n;i++){
              st=new StringTokenizer(br.readLine());
              for(int j=0;j<m;j++){
                  a[i][j]=Integer.parseInt(st.nextToken());
              }
          }
          return a;
      }
  }
  
  
  