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.lang.*;
import java.io.*;


public class Main
{   
    static int h[];
    static int dp[];
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner scan = new Scanner(System.in);
		int sz = scan.nextInt();
		int k = scan.nextInt();
		h = new int[sz];
		dp = new int[sz];
		for(int i = 0; i < sz; i++) h[i] = scan.nextInt();
		dp[0] = 0;
		dp[1] = Math.abs(h[1] - h[0]);
		for(int i = 2; i < sz; i++){
		    int min = Integer.MAX_VALUE;
		    for(int j = 1; i - j >= 0 && j <= k; j++)
		    min = Math.min(dp[i - j] + Math.abs(h[i] - h[i - j]), min);
		    dp[i] = min;
		}
		System.out.println(dp[sz - 1]);
	}
}