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.*;
import java.util.stream.IntStream;
class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
	int N = sc.nextInt();
    int K = sc.nextInt();
    int[] a = new int[N];
    for (int i = 0; i < N; i++) a[i] = sc.nextInt();
    int[] dp = new int[N];
    int b = K;
    for (int i = 1; i < N; i++) {
      if (i <= K) b--;
      for (int j = 1; j <= K-b; j++) {
        if (j == 1) {
          dp[i] = dp[i-j] + Math.abs(a[i] - a[i-j]);
        } else {
          dp[i] = Math.min(dp[i], dp[i-j] + Math.abs(a[i] - a[i-j]));
        }
      }
    }
    System.out.println(dp[N-1]);
  }
}