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 arr[]=new int[n];
      	for(int i=0;i<n;i++) arr[i]=sc.nextInt();
      
      	long dp[]=new long[n];
      	dp[0]=0;
      	dp[1]=Math.abs(arr[1]-arr[0]);
      	for(int i=2;i<n;i++){
          	long min=Long.MAX_VALUE;
        	for(int j=i-1;j>=0&&j>=i-k;j--){
            	long temp=dp[j]+Math.abs(arr[i]-arr[j]);
              	if(temp<min) min=temp;
            }
         	dp[i]=min;
        }
      	System.out.println(dp[n-1]);
    }
}