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[]){
   int N=0,k=0;

   Scanner obj=new Scanner(System.in);
   N=obj.nextInt();
k=obj.nextInt();
   int h[]=new int[N];
   for(int i=0;i<N;i++){
    h[i]=obj.nextInt();
   }
   int dp[]=new int[N];
	if(N==0)
      {
System.out.println(0);
return;
}
   if(N==1)
{     	
System.out.println(0);
return;
}
   if(N<=k+1)
     {
System.out.println(Math.abs(h[0]-h[N-1]));
return;
}
   dp[0]=0;
 //  dp[1]=Math.abs(h[0]-h[1]);
   for(int i=1;i<N;i++){
dp[i]=Integer.MAX_VALUE;
   for(int j =1;j<=i && j <=k;j++){
    dp[i]= Math.min(Math.abs(h[i]-h[i-j])+dp[i-j],dp[i]);
}
   }
   System.out.println(dp[N-1]);
 }
}
