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
/**
 * 
 */
//package AtCoder;
import java.util.*;
/**
 * @author Shivansh Singh
 *
 */
public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc=new Scanner(System.in);
		int n,k,i,min,j,p;
		n=sc.nextInt();
		k=sc.nextInt();
		int []arr=new int[n];
		for (i=0;i<n;i++)
			arr[i]=sc.nextInt();
		int []dp=new int[n+1];
		dp[0]=dp[1]=0;
		for (i=2;i<=n;i++) {
			j=i-1;
			min=dp[i-1]+(int)Math.abs(arr[i-1]-arr[i-2]);
			while(j>=1 && (i-j)<=k) {
				p=dp[j]+(int)Math.abs(arr[i-1]-arr[j-1]);
				min=(int)Math.min(min,p);
				j--;
			}
			dp[i]=min;
		}
		System.out.println(dp[n]);
		sc.close();
	}

}
