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 {
	static int n,k;
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		n=scan.nextInt();
		k=scan.nextInt();
		a=new int[n];
		for(int i=0;i<n;i++) a[i]=scan.nextInt();
		dp=new int[n];
		Arrays.fill(dp,-1);
		System.out.println(go(0));
	}
	static int[] dp;
	static int[] a;
	
	public static int go(int at) {
		if(at==a.length-1) return 0;
		if(dp[at]!=-1) return dp[at];
		int res=Integer.MAX_VALUE/2-5;
		for(int nxt=at+1;nxt<a.length&&nxt<=at+k;nxt++) {
			res=Math.min(res,Math.abs(a[at]-a[nxt])+go(nxt));
		}
		return dp[at]=res;
	}
}