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;
  import java.util.Arrays;
  
  public class main {
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
  
          int t = sc.nextInt();
          while (t > 0) {
              long l = sc.nextLong();
              long r = sc.nextLong();
              long thuong = 0;
              long mod = 0;
              long tong = r - l;
              if (tong == 0) {
                  System.out.println(0);
                  t--;
                  continue;
              }
  
              thuong = r / 2;
              if (thuong > l) {
                  mod = r % (thuong + 1);
                  System.out.println(mod);
                  t--;
                  continue;
              }
              if (thuong == l) {
                  mod = r % (l + 1);
              }
              if (thuong < l) {
  
                  mod = r % l;
  
              }
  
              System.out.println(mod);
              t--;
          }
      }
  
  }
  