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.*;
import java.io.*;

public class Main {
	static int dp[];
	static int k;
	public static void main(String[] args) throws IOException {	
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		k=s.nextInt();
		int a[]=new int[n];
		for(int i=0;i<n;i++)
			a[i]=s.nextInt();
		dp=new int[n];
		Arrays.fill(dp, -1);
		
		System.out.println(rec(0,a));
		
	}

	private static int rec(int i, int[] a) {
		if(dp[i]!=-1)
			return dp[i];
		if(i>=a.length)
			return (int)(1e5);
		if(i==a.length-1)
			return 0;
		
		
		int op=Integer.MAX_VALUE;
		for(int j=1;j<=k;j++) {
			if(i+j<a.length)
				op=Math.min(op,Math.abs(a[i]-a[i+j])+rec(i+j, a));
		}
		
		
		
		return dp[i]=op;
	}
}