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 A {
      public static void main(String[] args) throws IOException {
          MyScanner sc = new MyScanner();
          int T = sc.nextInt();
          while(T-- > 0) {
              int l = sc.nextInt();
              int r = sc.nextInt();
              if(l == r) {
                  System.out.println(0); 
                  continue;
              }
              l = Math.max((int)Math.floor(r / 2) + 1, l);
              System.out.println(r % l);
          }
      }
  }
  
  class MyScanner {
      BufferedReader br;
      StringTokenizer tok;
      MyScanner() throws IOException {
          br = new BufferedReader(new InputStreamReader(System.in));
          tok = new StringTokenizer("");
      }
      public String next() throws IOException {
          while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
          return tok.nextToken();
      }
      public int nextInt() throws IOException {
          return Integer.parseInt(next());
      }
  }
  
  