// Original C++ solution (slower):
/*

	Author: Luis Manuel D�az Bar�n (LUISMO)

	Problem:

	Online Judge:

	Idea:

*/

#include<bits/stdc++.h>

// Types

#define ll long long

#define ull unsigned long long

// IO

#define sf scanf

#define pf printf

#define mkp make_pair

#define fi first

#define se second

#define endl "\n"



using namespace std;



typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

const ll inf = 1e16 + 3;

const int mod = 1e9 + 7;

const int lim = 2e5 + 2;



int N, K;

ll hi[lim];



ll dp[lim];

int mark[lim];



ll DynamicProgramming(int idx)

{

    if(idx <= 0)

        return 0;



    if(mark[idx])

        return dp[idx];



    mark[idx] = true;



    DynamicProgramming(idx - 1);

    DynamicProgramming(idx - 2);





    int kidx =max(0,idx - K);

    ll aux = inf;

    for(int i = kidx; i < idx; i++)

    {

        ll curr = dp[i] + abs(hi[i] - hi[idx]);

        aux = min(aux, curr);

    }



    dp[idx] = aux;

    return dp[idx];

}





void solve()

{

    cin >> N >> K;

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

        cin >> hi[i];   



    ll answ = DynamicProgramming(N - 1);

    cout << answ;

}



void fastIO()

{

	cin.sync_with_stdio(false);

	cin.tie(0);

}



void IO()

{

	if(fopen("d:\\lmo.in","r") != NULL)

	{

		freopen("d:\\lmo.in","r",stdin);

	}

	else if(fopen("/media/Beijing/lmo.in","r") != NULL)

	{

		freopen("/media/Beijing/lmo.in", "r", stdin);

	}

}



int main()

{

	IO();



	fastIO();



	solve();

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

using namespace std;

#define MAX(a,b) (a>b?a:b)

#define MOD(a) (a>0?a:-(a))

#define custoate(i)

int main(){

  int npedras, maxpulo;

  scanf("%d %d",&npedras,&maxpulo);

  int hpedra[npedras],custoate[npedras];

  for (size_t i = 0; i < npedras; i++)

    scanf("%d", &hpedra[i]);

  custoate[0] = 0;

  int tempcost = 2147483647,k,tempcost2;

  for (size_t i = 1; i < npedras; i++,tempcost = 2147483647) {

    // printf("i = %d, tempcost = %d\n", i,tempcost);

    for (k= (maxpulo>i ? i : maxpulo); k > 0; k--) {

      // printf("k = %d ",k);

      tempcost2 = custoate[i-k] + MOD(hpedra[i-k]-hpedra[i]);

      if (tempcost > tempcost2)

        tempcost = tempcost2;

    }

    // printf("custo[%d] = %d",i,tempcost);

    custoate[i] = tempcost;

    // printf("\n\n");

  }

  printf("%d\n", custoate[npedras-1]);



}

// 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;





}