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.*;
  
  public class A {
    public static void main(String[] args) throws IOException {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  
      int t = Integer.parseInt(br.readLine());
      for (int i = 0; i < t; i++) {
        String[] inputs = br.readLine().split(" ");
        int l = Integer.parseInt(inputs[0]);
        int r = Integer.parseInt(inputs[1]);
        // r >= a >= b >= l
  
        int mod2 = (r+2) % 2;
        int newR = r + mod2;
        int halfR = newR/2;
        if (l <= halfR) {
          pw.println(halfR - 1);
        } else {
          pw.println(r % l);
        }
      }
  
      pw.flush();
    }
  }
  