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.util.StringTokenizer;
  
  
  public class Main {
  
      public static void main(String[] args) throws IOException {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          for (int i = 0; i < t; i++) {
              int l = sc.nextInt(), r = sc.nextInt() ;
              int a = r;
              int b = (int)Math.ceil((r+1)/2.0);
              pw.println(b >= l ? a%b: r-l);
              pw.flush();
         }
  
  
      }
  
  
  
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
  
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          public int[] nextIntArray(int n) throws IOException {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextInt();
              }
  
              return a;
          }
  
  
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          public String nextLine() throws IOException {
              return br.readLine();
          }
  
          public double nextDouble() throws IOException {
              String x = next();
              StringBuilder sb = new StringBuilder("0");
              double res = 0, f = 1;
              boolean dec = false, neg = false;
              int start = 0;
              if (x.charAt(0) == '-') {
                  neg = true;
                  start++;
              }
              for (int i = start; i < x.length(); i++)
                  if (x.charAt(i) == '.') {
                      res = Long.parseLong(sb.toString());
                      sb = new StringBuilder("0");
                      dec = true;
                  } else {
                      sb.append(x.charAt(i));
                      if (dec)
                          f *= 10;
                  }
              res += Long.parseLong(sb.toString()) / f;
              return res * (neg ? -1 : 1);
          }
  
          public boolean ready() throws IOException {
              return br.ready();
          }
  
  
      }
  }
  