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.Arrays;
import java.util.Scanner;
 
public class Main {
	public static void main (String[] args) {
      Scanner in = new Scanner(System.in);
      int N = in.nextInt();
      int K = in.nextInt();
      int[] a = new int[N];
      for (int i = 0; i < N; i++) {
        a[i] = in.nextInt();
      }
      int[] dp = new int[N];
      dp[0] = 0;
      dp[1] = Math.abs(a[0] - a[1]);
      for(int i = 2; i< N ;i++){
        int k = 1;
        int temp = Integer.MAX_VALUE;
        while(k <= K && i-k >=0){
          temp = Math.min(temp, dp[i-k] + Math.abs(a[i] - a[i-k]));
          k++;
        }
        dp[i] = temp;
      }
      System.out.println(dp[N-1]);
	}
}
