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
#include<cstdio>
#include<cmath>
#include<algorithm> 
using namespace std;
int n,k,a[100010],f[100010];
int main(){
  scanf("%d %d",&n,&k);
  for(int i=1;i<=n;i++) scanf("%d",&a[i]);
  for(int i=2;i<=n;i++){
    f[i]=2e9;
    for(int j=i-1;j>=1&&i-j<=k;j--){
    	f[i]=min(f[i],f[j]+abs(a[j]-a[i]));
	}
  }
  printf("%d",f[n]);
}