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;
int main()
{
	int n,k;
	cin>>n>>k;
	vector<long int>v(n);
	for(int i=0;i<n;i++)
	{
		cin>>v[i];
	}
	vector<long int>dp(n,INT_MAX);
	dp[0]=0;
	for(int i=0;i<n;i++)
	{
		for(int j=i+1;j<=i+k;j++)
		{
			if(j<n)
			dp[j]=min(dp[j],dp[i]+abs(v[i]-v[j]));
		}
	}
	cout<<dp[n-1]<<endl;
}