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.*;

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 n=0;n<N;n++){
			h[n] = sc.nextInt();
		}

		int dp[] = new int[N];
		dp[0] = 0;
		dp[1] = Math.abs(h[1]-h[0]);
		for(int n=2;n<N;n++){
			
			int tmp = -1;
			for(int k=1;k<=K;k++){

				if(n < k) break;

				int val = dp[n-k]+Math.abs(h[n]-h[n-k]);
				
				if(tmp==-1 || tmp > val)
				tmp = val;
			}
			dp[n] = tmp;
		}

		System.out.println(dp[N-1]);
	}
}