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 scan=new Scanner(System.in);
	int n=scan.nextInt();
	int k=scan.nextInt();
	int h[]=new int[n];
	for(int i=0;i<n;i++)
	{
		h[i]=scan.nextInt();
	}
	long dp[]=new long[n];
	Arrays.fill(dp,Long.MAX_VALUE);
	dp[0]=0;
	
	for(int i=1;i<n;i++)
	{
		//1 step to k steps
		long dist=-1;
		for(int cur=1;cur<=k;cur++) {
			if(i-cur<0)
			{
				break;
			}
			else {
			dp[i]=Math.min(Math.abs(h[i]-h[i-cur])+dp[i-cur],dp[i]);
			}
		}
//		System.out.println(dp[i]);
	}
	System.out.println(dp[n-1]);
	
}
}
