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: /******************************************************************************
  
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  import java.util.*;
  
  
  public class Main
  {
  	public static void main(String[] args) {
  	    Scanner sc=new Scanner(System.in);
  	    int t=sc.nextInt();
  	    while(t--!=0)
  	    {
  	        long a=sc.nextLong();
  	        long b=sc.nextLong();
  	        long x1=Math.max(a,b);
  	        long y1=Math.min(a,b);
  	        long x=(x1/2);
  	        //System.out.println(x);
  	        if(y1<=x)
  	        {
  	            System.out.println(x1%(x+1));
  	        }
  	        else
  	        {
  	            
  	            System.out.println(x1%y1);
  	        }
  	    }
  	}
  }
  