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 <bits/stdc++.h>
using namespace std;
#define int int64_t
#define MAX 100006
int a[MAX],dp[MAX];
signed main()
{
	int n,i,k,j;
	cin>>n>>k;
	for(i=1;i<=n;i++)
	{
	cin>>a[i];
	dp[i]=10000000000;
}
	dp[1]=0;
	for(i=2;i<=n;i++)
	for(j=1;j<=min(k,i-1);j++)
	dp[i]=min(dp[i-j]+abs(a[i]-a[i-j]),dp[i]);
	cout<<dp[n]<<"\n";
}