// Original C++ solution (slower):
#include<set>

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

#define rep(i, a, b) for(__typeof(b) i = a; i <= (b); i++)

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 200;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;

ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int n, k;

int h[maxn], dp[maxn];



int main()

{

	n = read(), k = read();

	rep(i, 1, n) h[i] = read();

	memset(dp, 0x3f, sizeof dp);

	dp[1] = 0;

	rep(i, 1, n){

		rep(j, 1, k) dp[i+j] = qmin(dp[i+j], dp[i] + abs(h[i] - h[i+j]));

	}

	printf("%d\n", dp[n]);

	return 0;

}

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

using namespace std;

#define int long long int



int n; //number of stones

int k; //farthest the frog can jump

vector<int> h; //height of the stones

vector<int> dp; //memorize heck



main(){

	cin >> n >> k;



	dp.assign(n, INT_MAX);

	

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

		int x;

		cin >> x;

		h.push_back(x);

	}



	dp[0] = 0;

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

		for(int j=1; j<=k; j++){

			if(i-j>=0){

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

			}

		}

	}



	cout << dp[n-1] << "\n";

	

}

// 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;





}