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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N=sc.nextInt();
		int K=sc.nextInt();
		int h[] = new int[N];
		for(int i=0;i<N;i++) {
			h[i]=sc.nextInt();
		}
		int cost[] = new int[N];
		cost[N-1] = 0;
		cost[N-1-1] = Math.abs(h[N-1] - h[N-1-1]);
		
		for(int i=N-2-1;i>=0;i--) {
			ArrayList<Integer> list = new ArrayList<Integer>();
			for(int j=1;j<=Math.min(K,N-1-i);j++) {
				list.add(cost[i+j]+Math.abs(h[i+j]-h[i]));
			}
			Collections.sort(list);
			cost[i] = list.get(0);
		}
		
		System.out.println(cost[0]);
	}

}
