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.*;
  import java.text.*;
  import java.math.*;
  
  public class Main {
  	public static void main(String[] args) throws IOException{
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb = new StringBuilder();
  		int t = Integer.parseInt(br.readLine());
          while(t-- > 0){
              StringTokenizer st = new StringTokenizer(br.readLine());
              int l = Integer.parseInt(st.nextToken());
              int r = Integer.parseInt(st.nextToken());
              int res = r/2 + 1;
              if(res < l)
                  res = l;
              sb.append(r%res + "\n");
          }
  		System.out.print(sb);
  	}
  }
  
  
  