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.util.*;
  
  public class test {
  
  	public static void main(String[] args) {
  
  		Scanner sc = new Scanner(System.in);
  
  		int t = sc.nextInt();
  
  		while(t-- > 0) {
  
  			int l = sc.nextInt(); int r = sc.nextInt();
  
  			if((r - l) < r/2) {
  
  				System.out.println(r - l);
  
  			}
  
  			else {
  
  				if(r % 2 == 0) {
  
  					System.out.println((r/2) - 1);
  
  				}
  
  				else {
  
  					System.out.println(r/2);
  
  				}
  
  			}
  
  		}
  
  		sc.close();
  
  	}
  
  }