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.io.*;
  import java.math.BigInteger;
  import java.util.*;
  
  public class Main {
      public static void main(String[] args) throws IOException {
          //br = new BufferedReader(new FileReader("input.txt"));
          //out = new PrintWriter("output.txt");
          int t = nextInt();
          while (t-- != 0) {
              long l = nextLong();
              long r = nextLong();
             if(r/2+1<l)out.println(r%l);
             else out.println(r%(r/2+1));
          }
          out.close();
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  }
  
  class Ayo implements Comparable<Ayo> {
      long r;
      long k;
  
      Ayo(long r, long k) {
          this.r = r;
          this.k = k;
      }
  
      public int compareTo(Ayo o) {
          return Long.compare(r, o.r);
      }
  }