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.io.*;
import java.util.*;

public class MyClass {
	public static void main(String[] args)
	{
		// Dividend
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		
		while(t-->0)
		{
		
		long a = s.nextLong();

		// Divisor
		long b = s.nextLong();

		// Mod
		long ans=0;
	
		long di= (long)Math.floor(b/2)+1;
		if(di>=a)
		System.out.println(b%di);
	    else
	    {
		 
	 
		   System.out.println(b%a);
	     }
		}
	}
}