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.*;
  
  public class Demo{
      public static void main(String[] args) {
  
          Scanner sc = new Scanner(System.in);
  
          int t = sc.nextInt();        
          for(int i=0; i<t; i++){
              long l = sc.nextInt();
              long r = sc.nextInt();
              // Long[][] dp = new Long[Integer.MAX_VALUE][Integer.MAX_VALUE];
              System.out.println(solve(l, r));
          }
      }
      public static long solve(long l, long r){
          
          if(l> r/2){
              return r % l;
          }else{
              if(r%2 != 0){
                  r++;
              }
             long x = r/2 + 1;
              return r % x;
          }
  
          // if(dp[(int) l][(int) r] != null) return dp[(int) l][(int) r];
          // // System.out.println(l +"  "+ r + " mode "+ mod);
          // return dp[(int) l][(int) r] =  Math.max(mod,
          //              Math.max(
          //                 solve(l+1, r, r%l, dp),
          //                 solve(l, r-1, r%l, dp)
          //             )
          //         );
      }        
  }
  
  
  