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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] h = new int[n];
		for(int i = 0; i < n; i++) {
			h[i] = sc.nextInt();
		}
		int[] dp = new int[n];
		dp[0] = 0;
		for(int i = 1; i < n; i++) {
			dp[i] = Integer.MAX_VALUE;
		}
		for(int i = 1; i < n; i++) {
			for(int j = 1; j <= k; j++) {
				if(i - j >= 0) {
					dp[i] = Math.min(dp[i], dp[i - j] + Math.abs(h[i] - h[i - j]));
				}
			}
		}
		System.out.println(dp[n - 1]);
	}

}
