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.Scanner;
  
  /**
   *
   * @author Acer
   */
  public class NewClass {
      
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int T = sc.nextInt();
          while(T-- > 0){
              int l = sc.nextInt();
              int r = sc.nextInt();
              
              if(l == r){
                  System.out.println(0);
                  continue;
              }
              
              int x = r/2;
              if(l <= x){
                  System.out.println(r-(x+1));
              }
              else {
                  System.out.println(r-l);
              }
              
          }
      }
      
  }