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 sc=new Scanner(System.in);
	    int n=sc.nextInt();
	       int k=sc.nextInt();
	    int a[]=new int[n];
	    for(int i=0;i<n;i++){
	    a[i]=sc.nextInt();
	    }
	    int dp[]=new int[n];
	   
	    dp[0]=0;
	    dp[1]=Math.abs(a[1]-a[0]);
	    for(int i=2;i<n;i++){
	         //dp[i]=Integer.MAX_VALUE;
	         dp[i] = Integer.MAX_VALUE;
	      for(int j=0;j<=k;j++){
	          if(i>=j)
	    dp[i]=Math.min(dp[i-j]+Math.abs(a[i]-a[i-j]),dp[i]);
	      }
	    }
	  System.out.println(dp[n-1]);

	}

}
