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 scn = new Scanner(System.in);
		int n = scn.nextInt();
		int k = scn.nextInt();
		int arr[] = new int[n+1];
		for (int i = 1; i <=n; i++) {
			arr[i] = scn.nextInt();

		}
		
		System.out.println(frog(arr, k,n));

	}

	public static int frog(int[] arr, int k,int n) {
		int dp[] = new int[n + 1];
		
		for (int i = 2; i <=n; i++) {
			dp[i] = Integer.MAX_VALUE;
			for (int j = i - 1; j >= 1 && (i - j) <= k; j--) {
				dp[i] = Math.min(dp[i], Math.abs(arr[i] - arr[j]) + dp[j]);
			}

		}
		return dp[n];

	}
}