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.BufferedReader;
  import java.math.BigDecimal;
  import java.math.BigInteger;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  public class Main {
  	static int n,m;
  	static int[] A;
  	static int[][] B;
  	static long ans,cnt;
  
  	public static void main(String[] args)throws IOException {
  		// TODO Auto-generated method stub
  		Scanner in = new Scanner(System.in);
  		n = in.nextInt();
  		for(int x=0;x<n;x++) {
  			int a = in.nextInt();
  			m = in.nextInt();
  			//int a = in.nextInt();
  			int mid = 0;
  			int num = 0;
  			if(m%2==0) {
  				mid = m/2-1;
  				num = mid+1;
  			}
  			else {
  				mid = (m+1)/2-1;
  				num = mid+1;
  			}
  			if(num>=a) {
  				System.out.println(mid);
  			}
  			else {
  				System.out.println(m%a);
  			}
  			// 14 27 
  		}
  		
  	}
  }
  