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.io.*;
class Main{
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String inp[]=br.readLine().split(" ");
		int n = Integer.parseInt(inp[0]);
		int k = Integer.parseInt(inp[1]);
		String in[]=br.readLine().split(" ");
		int arr[]=new int[n];
		for(int i=0;i<n;i++)
			arr[i]=Integer.parseInt(in[i]);
		int dp[]=new int[n];
		Arrays.fill(dp,1000000005);	
		dp[0]=0;	
		for(int i=0;i<n;i++){
			for(int j=1;j<=k;j++){
				if(i+j <n)
					dp[i+j]=Math.min(dp[i+j],dp[i]+Math.abs(arr[i]-arr[i+j]));
			}	

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