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 input  = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		int arr[] = new int[n];
		for(int i=0;i<n;i++) {
			arr[i] = input.nextInt();
		}
		System.out.println(func(n,arr,k));
	}
	static int func(int n,int arr[],int k) {
		int dp[] = new int[n];
		dp[0] = 0;
		for(int i=1;i<n;i++) {
			int ans = Integer.MAX_VALUE;
			for(int j=1;j<=k;j++) {
				if(i-j>=0) {
					ans = Math.min(ans,Math.abs(arr[i]-arr[i-j])+dp[i-j]);
				}
			}
			dp[i] = ans;
		}
		return dp[n-1];
	}
}
