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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author zipM
   */
  public class Main {
    public static void main(String[] args) {
      InputStream inputStream = System.in;
      OutputStream outputStream = System.out;
      InputReader in = new InputReader(inputStream);
      PrintWriter out = new PrintWriter(outputStream);
      TaskA solver = new TaskA();
      solver.solve(1, in, out);
      out.close();
    }
  
    static class TaskA {
      public void solve(int testNumber, InputReader in, PrintWriter out) {
        int t = in.nextInt();
        while (t-- > 0) {
          int l = in.nextInt();
          int r = in.nextInt();
          int R = r;
          int ans = r % l;
          while (l <= r) {
            int h = (l + r) / 2;
            if (2 * h - 1 <= R) {
              ans = Math.max(ans, h - 1);
              l = h + 1;
            } else {
              r = h - 1;
            }
          }
          out.println(ans);
        }
      }
  
    }
  
    static class InputReader {
      public BufferedReader reader;
      public StringTokenizer tokenizer;
  
      public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 65536 / 2);
        tokenizer = null;
      }
  
      public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
          try {
            tokenizer = new StringTokenizer(reader.readLine());
          } catch (IOException e) {
            throw new RuntimeException(e);
          }
        }
        return tokenizer.nextToken();
      }
  
      public int nextInt() {
        return Integer.parseInt(next());
      }
  
    }
  }
  
  