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 sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    int[] h = new int[n];
    int[] DP = new int[n];
    Arrays.fill(DP,Integer.MAX_VALUE);

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

    DP[0] = 0;
    DP[1] = Math.abs(h[1] -h[0]);

    for(int i=2; i<n; i++){
      for(int j=1; j<=k;j++){
        DP[i] = Math.min(DP[i],DP[i-j]+Math.abs(h[i]-h[i-j]));
        if(i-j==0){break;}
      }

    }

    System.out.println(DP[n-1]);

  }
}