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.*;
public class ayush {
   public static void main(String[] args) {
       try{
            Scanner sc = new Scanner(System.in);
            int t= sc.nextInt();
            while(t>0){
                int l,r,rem=0;
                l=sc.nextInt();
                r=sc.nextInt();
                int num1=r%l;
                int num2=(r+1)/2;
                if(num2<l){
                    num2=0;
                }
                else{
                    num2=num2-1;
                }
                System.out.println(Math.max(num1,num2));
                
            }
            t--;
            }catch(Exception e){
                return;
            }
 }
}
