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 cost[] =new int[N];
		for(int i=0;i<N;i++)
		{
		
			cost[i]= sc.nextInt();
		}
		int dp[] =new int [N+1];
		Arrays.fill(dp,Integer.MAX_VALUE);
		dp[0]=0;
		dp[1]=Math.abs(cost[1]-cost[0]);
		//dp[1]=cost[1]-cost[0];
		
		for(int i=2;i<N;i++)
		{
			 for(int j=i-1,jump=0;j>=0 && jump<K;j--,jump++)
			 {
			 	dp[i]=Math.min(dp[i],Math.abs(cost[i]-cost[j]) +dp[j]);
			 }
		}
		
		System.out.println(dp[N-1]);
	}

}