// Original C++ solution (slower):
#include <bits/stdc++.h>

//freopen("transmitters.in","r",stdin);

//cout<<fixed<<setprecision(6)<<n<<endl;

// transform(s.begin(), s.end(), s.begin(), ::tolower);

//double PI=3.14159265359,c=0;

#define all(v) v.begin(),v.end()

#define endl '\n'

#define ll long long

//#define f(n) for(int i=;i<n;i++)

double PI = acos(-1);

using namespace std;

void fast()

{

    std::ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}

ll arr[10000000],n,k;

vector<ll>v(1e6,-1);

ll minCost(ll indx)

{

    ll path1=LLONG_MAX;

    if(indx==n-1)

        return 0;

        

    if(v[indx]!=-1)

        return v[indx];

    for(int i=0;i<k;i++)

    {

      if((indx+1+i)<n)

          path1=min(path1,(abs(arr[indx]-arr[indx+1+i])+minCost(indx+1+i)));

    }

    return v[indx]=path1;

}

int main()

{

    fast();

    cin>>n>>k;

    for(int i=0; i<n; i++)

        cin>>arr[i];



    cout<<minCost(0)<<endl;



    return 0;

}


// Optimized solution to the same problem (faster):
/*

				" اللهم صلي و سلم علي سيدنا محمد و علي اله وصحبه اجمعين "

*/





#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#define f for

#define ll long long

#define ull unsigned long long

#define endl "\n"

using namespace std;

#define BI  acos(-1)

#define all(a) a.begin(), a.end()

void Open()

{

#ifndef ONLINE_JUDGE

	freopen("Input.txt", "r", stdin);

	freopen("Output.txt", "w", stdout);

#endif

}

ll gcd(ll a, ll b)

{

	return b == 0 ? a : gcd(b, a % b);

}

void madesetprobblem()

{

	//$ $

	//less than or equal => $ 1 \leq 100 $

	//  $(1 \le A,B \le 10^5 )$

	// a[i] => $a_i$

	// a[i][j] => $a_{i,j}$

	// double qoutes ''yext''

	/*

		image

		\begine{center}

		\includegraphics{image.jpg}

		\end{center}

	*/

	/*

		list

		\begin{itemize}

			\item first one

			\item second one

		\end{itemize}

	*/

}

int n,k;

vector<int>a(1e5 + 2),dp(1e5+2,-1);

int ans(int c = 0)

{

	if (c == n-1)

		return 0;

	int ans1 = INT_MAX, ans2 = INT_MAX;

	if (dp[c] != -1)

		ans1 = dp[c];

	else

	{

		f(int i = 1; i <= k; i++)

		{

			if (c + i < n)

				ans1 = min(ans1,(abs(a[c] - a[c + i]) + ans(c + i)));

		}

		dp[c] = ans1;

	}

	//cout <<c<<" "<< ans1 << " " << ans2 << endl;

	return ans1	;

}

int main()

{

	// توكلت علي الله 

//Open();

	ios::sync_with_stdio(false);

	cin.tie(NULL);

	cout.tie(NULL);

	cin >> n>>k;

	f(int i = 0; i < n; i++)

	{

		cin >> a[i];

	}

	cout << ans();

}

// Fastest known solution for this problem:
#include <iostream>

#include <array>

#include <vector>

#include <iomanip>

#include <string>

#include <cmath>

#include <array>

#include <algorithm>

#include <queue>

#include <map>

using namespace std;



typedef  pair<int,int> P;

typedef long long LL;



const LL  INF=1<<31;







int main(){

     int n,k;

     cin>>n>>k;

     int h[100000];

     for(int i=0;i<n;++i){

          cin>>h[i];

     }

     LL dp[100001];



      for(int i=0;i<n;++i){

           dp[i]=INF;

      }

            dp[0]=0;

            dp[1]=abs(h[0]-h[1]);







     for(int i=2;i<n;++i){

          dp[i]=dp[i-1]+abs(h[i-1]-h[i]);

          for(int j=2;j<=k;++j){



               if(j<=i)dp[i]=min(dp[i],dp[i-j]+abs(h[i-j]-h[i]));

          }

     }     



     cout<<dp[n-1]<<endl;





}