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: // Don't place your source in a package
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  // Please name your class Main
  public class Main {
      
      private static final FastReader input = new FastReader(); 
      
  	public static void main (String[] args) throws java.lang.Exception {
  
        int t = input.nextInt();
    
          while(t-- != 0){
              int l = input.nextInt();
              int r = input.nextInt();
  
              if(r < l * 2){
                  System.out.println(r - l);
              } else{
                  System.out.println((r - 1) / 2);
              }
          }
     }
  }
  
  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) {
                  throw new RuntimeException(e);
              }
          }
          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) {
              throw new RuntimeException(e);
          }
          return str;
      }
  }
  
  