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.
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main 
{
        public static void main(String... args) throws IOException
        {
                int tt;
                //Scanner sc= new Scanner(System.in);
          BufferedReader inf=new BufferedReader(new InputStreamReader(System.in));      
          StringTokenizer st=new StringTokenizer(inf.readLine());      
                tt=Integer.parseInt(st.nextToken());
                int l,r;
                 while((tt--)>0)
                 { 
                st=new StringTokenizer(inf.readLine());
                 l=Integer.parseInt(st.nextToken());
                  r=Integer.parseInt(st.nextToken());
                //l=Integer.parseInt(inf.readLine());
                //r=Integer.parseInt(inf.readLine());
                System.out.println((((r-1)/2 +1)>=l)?((r-1)/2):(r%l));  
                 }
        }
}