There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.

There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:

* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.



Find the minimum possible total cost incurred before the frog reaches Stone N.

Constraints

* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4

Input

Input is given from Standard Input in the following format:


N K
h_1 h_2 \ldots h_N


Output

Print the minimum possible total cost incurred.

Examples

Input

5 3
10 30 40 50 20


Output

30


Input

3 1
10 20 10


Output

20


Input

2 100
10 10


Output

0


Input

10 4
40 10 20 70 80 10 20 70 80 60


Output

40
import java.util.*;
public class Main
{
	public static void main(String[] args)
	{
	Scanner s=new Scanner(System.in);
	int n=s.nextInt();
	int h[]=new int[n];
 	int k=s.nextInt();
	for(int i=0;i<n;i++)
	h[i]=s.nextInt();
	int dp[]=new int[n];
	dp[0]=0;
	int min=Integer.MAX_VALUE;
	int a1=0;
	for(int i=1;i<n;i++)
	{
	 if(i<=k)
	dp[i]=Math.abs(h[i]-h[0]);
	  else
	   {
	      
	    for(int j=i-k;j<i;j++)
		{
		if(dp[j]+Math.abs(h[j]-h[i])<min)
		min=dp[j]+Math.abs(h[i]-h[j]);
		
		}
	dp[i]=min;
	min=Integer.MAX_VALUE;
	   }
	}
	
	System.out.println(dp[n-1]);
	}
}