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.*;
  import java.util.*;
  import java.lang.*;
  import static java.lang.Math.*;
  
  // Sachin_2961 submission //
  public class Codeforces {
  
      public void solve() {
          long l = fs.nLong(), r = fs.nLong();
          if( l == r){
              out.println("0");
              return;
          }
          if(2*l > r ){
              long ans = r%l;
              out.println(ans);
              return;
          }
          long ans = r/2 - (r%2==0?1:0);
          out.println(ans);
      }
      static boolean multipleTestCase = true; static FastScanner fs; static PrintWriter out;
      public void run(){
          fs = new FastScanner();
          out = new PrintWriter(System.out);
  
          int tc = (multipleTestCase)?fs.nInt():1;
          while (tc-->0)solve();
          out.flush();
          out.close();
      }
      public static void main(String[]args){
          try{
              new Codeforces().run();
          }catch (Exception e){
              e.printStackTrace();
          }
  
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String n() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
          String Line()
          {
              String str = "";
              try
              {
                  str = br.readLine();
              }catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
          int nInt() {return Integer.parseInt(n()); }
          long nLong() {return Long.parseLong(n());}
          double nDouble(){return Double.parseDouble(n());}
          int[]aI(int n){
              int[]ar = new int[n];
              for(int i=0;i<n;i++)
                  ar[i] = nInt();
              return ar;
          }
      }
      public static void sort(int[] arr){
          ArrayList<Integer> ls = new ArrayList<Integer>();
          for(int x: arr)
              ls.add(x);
          Collections.sort(ls);
          for(int i=0; i < arr.length; i++)
              arr[i] = ls.get(i);
      }
  }