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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main (String[] args) {
		Scanner s=new Scanner(System.in);    
		int N=s.nextInt();		
		int K=s.nextInt();	
		int h[]=new int[N];
		for(int i=0;i<N;i++)
		{
			h[i]=s.nextInt();
		}
		int dp[]=new int[N];
		Arrays.fill(dp,-1);
		dp[0]=0;
		int i=0,d=0;
		while(i<N-1)
		{
			int j=1;
			while(j<=K)
			{
				if(i+j<N)
				{
					d=Math.abs(h[i]-h[i+j])+dp[i];
					if(dp[i+j]==-1)
						dp[i+j]=d;
					else
						if(dp[i+j]>d)
							dp[i+j]=d;
				}
				else
					break;
				j+=1;
			}
			i+=1;
		}
		System.out.print(dp[N-1]);

	}
}
