stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int n;

int h[100005];

int dp[100005];



int solve(int x){

    if (x >= n-1) return 0;

    else if (dp[x] != -1) return dp[x];

    else if (x == n-2){

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

    }

    else {

        return dp[x] = min(abs(h[x+1]-h[x])+solve(x+1),abs(h[x+2]-h[x])+solve(x+2));

    }

}



int main(int argc, char const *argv[]) {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin >> n;

    memset(&dp,-1,sizeof(dp));

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

        cin >> h[i];

    }

    cout << solve(0) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4141288480909098x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define inf INFINITY

#define file freopen( "input.txt" , "r" , stdin) ; freopen( "output.txt" , "w" , stdout) ;

#define Gamed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define _3102_ main()

#define endl '\n'

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

using namespace std;

const int N = 1e5 + 100;

const ll MX = 1e9;

const ll mod = 1e9 + 7;

const double eps = 1e-9 ;



ll n, h[N], dp[N] ;

ll solve( int i ){

	if ( i < 0 ) return 0 ;

	ll &ret = dp[i] ;

	if ( ~ret ) return ret ;

	if ( i == 0 ) return ret = 0 ;

	if ( i == 1 ) return ret = abs( h[1] - h[0]) ; 

	return ret = min ( solve(i - 1 ) + abs(h[i] - h[i-1]) , solve ( i - 2 ) + abs ( h[i] - h[i-2])) ;

}



int _3102_{

	Gamed

	cin >> n ;

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

		cin >> h[i] ;

	}

	memset ( dp , -1 , sizeof dp ) ;

	cout << solve ( n - 1 ) ;



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8156114145493796x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>



#define fst first

#define snd second

#define fore(i,a,b) for(int i=a, ThxDem=b; i<ThxDem; ++i)

#define pb push_back

#define ALL(s) s.begin(),s.end()

#define FIN ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

#define SZ(s) int(s.size())

#define DBG(x) cerr <<#x << " = " << (x) << "\n"



#define SQR(x) ((x) * (x))



#define PI 3.1415926535897932385

#define INF 1000111222

#define eps 1e-7

#define maxN 11



using namespace std;



typedef long long ll;

typedef vector<ll> vll;

typedef pair<ll,ll> pll;

typedef pair<int,int> pii;

typedef vector<string> vstr;

typedef vector<vll> vvll;

typedef vector<bool> vbool;

typedef vector<pll> vpll;

typedef vector<vpll> vvpll;

typedef vector<vbool> vvbool;



vll ns,dp;

ll n;



ll solve(ll pos){

	ll & d = dp[pos];

	if(d != -1) return d;

	ll v=INF;

	if((pos+1)<n)v = min(v,abs(ns[pos+1]-ns[pos])+solve(pos+1));

	if((pos+2)<n)v = min(v,abs(ns[pos+2]-ns[pos])+solve(pos+2));

	if(pos == (n-1)) v = 0;

	return d = v;

}



int main(){FIN

	

	cin >> n;

	ns.resize(n);

	dp.assign(100010,-1);

	fore(i,0,n)cin >> ns[i];

	

	solve(0);

	

	cout << dp[0] << "\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8267123777796517x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



#define fi first

#define se second

#define pb push_back

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

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



typedef long long ll;

typedef long double ld;

typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

typedef vector<int> vi;

typedef vector<vi> vvi;

typedef vector<pii> vii;



const int MAX = 1e5;

const int INF = 1e9;

vi stone(MAX + 5);

vi dp(MAX + 5, -1);

int n;



int sol(int i){

    if(i >= n) return INF;

    if(i == n - 1) return 0;

    if(dp[i] != -1) return dp[i];

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

    if(i + 2 < n){

        dp[i] = min(dp[i], sol(i + 2) + abs(stone[i] - stone[i + 2]));

    }

    return dp[i];

}



int main(){

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    #ifdef LOCAL

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

    #else

        #define endl '\n'

    #endif

    cin >> n;

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

        cin >> stone[i];

    cout << sol(0) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8336092934114858x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

//-------------------------------------------MACROS---------------------------------------------

#define 	TIEMPO_CORRIDA		cerr << "tiempo: "<<1.0*clock()/CLOCKS_PER_SEC <<" seg.\n";

#define 	FILES				freopen("input.txt","r",stdin),freopen("output.txt","w",stdout);

#define 	FORIT(it,e)			for(it = e.begin() ; it != e.end() ; it++)

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

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

#define 	FORSW(i,a,b,sw)		for(i = a ; i < b && sw ; i++)

#define 	RFORSW(i,a,b,sw)	for(i = a ; i >= b && sw ; i--)

#define 	fast 				ios::sync_with_stdio(false),cin.tie(),cout.tie();

#define 	to_int(x) 			atoi(x.data())

#define 	to_long(x) 			atoll(x.data())

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

#define 	RALL(x) 			x.rbegin(),x.rend()

typedef 	long long 			ll;

typedef 	pair<int,int> 		pi;

typedef 	pair<ll,ll> 		pl;

const		int 				modN = 1e9+7;

const		double 				PI = 3.1415926535897932384626433832795028841;

//---------------------------------------------------------------------------------------------

const int tam = 6e5;

const int oo=1e9;

int ar[tam],memo[tam];

int n;



int f(int x){

	if (memo[x]!=-1) return memo[x];

	if(x == n-1) return memo[x]=0;

	int a=oo , b=oo;

	if(x+1 <n) a = f(x+1) + abs(ar[x+1]-ar[x]); 

	if(x+2 <n) b = f(x+2) + abs(ar[x+2]-ar[x]);

	return memo[x]=min(a,b);

}



int main(){

	int i;

	while(scanf("%d",&n)==1){

		memset(memo,-1,sizeof memo);

		FOR(i,0,n) scanf("%d",&ar[i]);

		printf("%d\n",f(0));

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.39062880752921x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

const int maxn=100000+10;

int n,h[maxn],dp[maxn];

int main()

{

	cin>>n;

	for(int i=1;i<=n;i++)cin>>h[i];

	memset(dp,0x3f3f3f3f,sizeof(dp));

	dp[1]=0;

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

	{

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

		dp[i]=min(dp[i],dp[i-2]+abs(h[i]-h[i-2]));

	}

	cout<<dp[n]<<endl;

	return 0;

}

//Phoenix-ZH

// Faster C++ solution for the same problem (measured speedup: 1.4006762090600589x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>





using namespace std;

int n;

vector<int> vec;



int meme[100007];



int solve(int index){



    if(index==n-1){

        return 0;

    }

    if(meme[index]!=-1){

        return meme[index];

    }

    

    int ret1 = 100000000;

    int ret2 = 100000000;

    if(index+1<n){

         ret1 = abs (vec[index]- vec[index+1]) + solve(index+1); 

    }

    if(index+2<n){

         ret2 = abs (vec[index]- vec[index+2]) + solve(index+2); 

    }





    return meme[index] = min(ret1,ret2);

}



int main(){

    cin>>n;

    memset(meme,-1,sizeof(meme));

    int num;

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

        cin>>num;

        vec.push_back(num);

    }

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4033492617366312x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define ll long long

#define endl "\n"

using namespace std;

inline int D()

{

    int t;

    scanf("%d",&t);

    return t;

}

inline ll llD()

{

    ll t;

    scanf("%lld",&t);

    return t;

}



const int Max=1e6+9;

const int OO=-1e6;

int st[4]= {1,0,0,-1};

int nt[4]= {0,1,-1,0};

int n,m;

bool valid(int x,int y)

{

    return x>-1 && y>-1 && x<n && y<m;

}

ll mem[Max];

void init()

{

    memset(mem,-1,sizeof mem);

}

//////////////////////////////////////////////////////////



ll a[Max];



ll solve (int i )

{

    if(i==n-1)return 0;

    if(i>=n)

        return Max;



    ll &ret=mem[i];

    if(~ret)return ret;



    ll take1 =solve(i+1)+abs(a[i]-a[i+1]);

    ll take2 =solve(i+2)+abs(a[i]-a[i+2]);



     return mem[i]=min(take2,take1);

}



void Fast()

{

    std::ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    cerr.tie(0);

}

int main()

{

   init();

   n =D();

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

   {

       a[i]=llD();

   }

   printf("%lld",solve(0));



}


// Faster C++ solution for the same problem (measured speedup: 9.379601678333586x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



typedef long long ll;

const double eps=1e-6,pi=acos(-1);

const int N=1e5+5,M=1e9+7,SEGM=4*N,OO=0x3f3f3f3f;

int n,m,a[N],b[N],c[N];

int mem[N];



int solve(int i=0){

    if(i>=n) return 1e9;

    if(i==n-1) return 0;

    int &ret=mem[i];

    if(~ret) return ret;

    ret=1e9;

    ret=min(ret,solve(i+1)+abs(a[i]-a[i+1]));

    ret=min(ret,solve(i+2)+abs(a[i]-a[i+2]));

    return ret;

}



int main(){

    //freopen("myfile.txt","w",stdout);

    scanf("%d",&n);

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

        scanf("%d",a+i);

    memset(mem,-1,sizeof mem);

    printf("%d",solve());

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.388095566636733x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

 

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

//#pragma GCC optimize("Ofast")

//1.0 * clock() / CLOCKS_PER_SEC

 

#define fi first

#define se second

#define ll long long int

#define dl double long

 

using namespace std;

 

const int N = 1e5 + 7;

const ll mod = 1e9 + 7;



vector<int> dp(N);



void solve1()

{

	int n;

	cin >> n;

	int ar[n];

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

		cin >> ar[i];

	dp[1] = 0;

	dp[2] = abs(ar[1] - ar[0]);

	for(int i = 3; i <= n; i++){

		dp[i] = min(abs(ar[i - 1] - ar[i - 2]) + dp[i - 1], abs(ar[i - 1] - ar[i - 3]) + dp[i - 2]);

	}

	cout << dp[n];

}

int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0); srand(time(0));

    //freopen( "input.txt" , "r" , stdin );

    //freopen( "output.txt" , "w" , stdout );

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

    //freopen( "blist.out" , "w" , stdout );

    

    int cghf = 1;//cin >> cghf;

    while( cghf-- ){

        solve1();

    }

}


// Faster C++ solution for the same problem (measured speedup: 1.4177069319258073x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include<bits/stdc++.h>



typedef   long long  int ll;

typedef  long double ld;

typedef std::vector<ll> vi;

typedef std::vector<std::vector<ll> > vv;



#define MOD 1000000007

#define IO ios_base::sync_with_stdio(false);cin.tie(NULL);

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

#define mp make_pair

#define pb(x) push_back(x)

#define fi first

#define se second

#define print(vec,a,b) for(ll i=a;i<b;i++) cout<<vec[i]<<" ";cout<<endl;

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

#define input(vec,a,b) for(ll i = a;i<b;i++) cin>>vec[i];

#define ms(a,val) memset(a,val,sizeof(a))

using namespace std;



const int N = 1e5 + 5;





void solve(){

            ll i,j,n,m;

            ll mincost[N];

            ms(mincost,0);

            cin>>n;

            ll a[n];

            input(a,0,n);

            mincost[1] = abs(a[0]-a[1]);//a[0];

            fo(i,2,n)

            {

				mincost[i] = min(mincost[i-1]+abs(a[i]-a[i-1]),mincost[i-2]+abs(a[i-2]-a[i]));

			}

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

            

    

            

						            



















}



int main()

{

    IO;

      ll t=1;

      //cin>>t;





      while(t--)

      {



          solve();

      }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8223134115539372x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
/* In the name of Allah */



#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define pii pair<int, int>

#define pll pair<long long, long long>

#define pb push_back

#define LL_INF 1LL << 62

#define INF 1 << 30

#define SetBit(x, k) (x |= (1LL << k))

#define ClearBit(x, k) (x &= ~(1LL << k))

#define CheckBit(x, k) (x & (1LL << k))

#define mod 1000000007



int n;

ll x[100005];

ll dp[100005];



ll solve(int pos){

  if(pos>=n-1) return 0;

  if(dp[pos]!=-1) return dp[pos];



  ll res = LL_INF;

  if(pos+1<n) res = min(res, abs(x[pos+1]-x[pos])+solve(pos+1));

  if(pos+2<n) res = min(res, abs(x[pos+2]-x[pos])+solve(pos+2));



  return dp[pos] = res;

}



int main()

{

  memset(dp, -1, sizeof dp);

  cin >> n;

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

    cin >> x[i];



  cout << solve(0) << "\n";



  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8063264327179394x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int dp[100005], a[100005];

int n;



int solve(int n)

{

    dp[1]=0;

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

    {

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

        if(i>2)

            dp[i]=min(dp[i], dp[i-2] + abs(a[i]-a[i-2]));

    }

    return dp[n];

    

}

int main()

{

    cin>>n;

    for(int i=1; i<=n; i++)

        cin>>a[i];

    cout<< solve(n) <<endl;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int n , a[1000000] , frog , memo[1000000];



int f(int n){

	if(n == 1)return 0;

	if(n == 2)return abs(a[1] - a[2]);

	if(memo[n] != -1)return memo[n];

	memo[n] = min(f(n-1) + abs(a[n] - a[n-1]) , f(n-2) + abs(a[n] - a[n-2]));

	return memo[n];

}

int main(){

	memset(memo , -1 , sizeof(memo));

	scanf("%d" , &n);

	for(int i = 1 ; i <= n ; i++){

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

	}

	printf("%d" , f(n));

}

// Faster C++ solution for the same problem (measured speedup: 1.200509789142155x):


#include<bits/stdc++.h>

#define ll long long int

#define s(l) scanf("%lld",&l)

#define ss(lc) scanf("%s",lc)



#define ppp push_back

#define p(x) printf("%lld\n",x)

#define sr string

#define vr vector

/*int ff(int w)

{

    if(w==1){return 1;}

    else if(w==0){return 1;}

    else{return w*ff(w-1);}

}*/

/*int pr(int ww)

{

    int xs=ww/2;

    int sx=1;

    for(int i=2;i<=xs;i++)

    {

        if(ww%i==0){sx=0;break;}



    }

    return sx;

}*/

ll a[200005],b[200005];

using namespace std;

int main()

{

    vr<ll> mn;

    memset(a,0,sizeof(a));

    memset(b,0,sizeof(b));

  ll x=0,y,z=0,zz,i,j,k=0,l,m,n,o,p,q,r,ss;

  s(n);

vr<ll>dp(n,0);

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

{

    s(a[i]);

}

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

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

{

    dp[i]=min(abs(a[i]-a[i-1])+dp[i-1],abs(a[i]-a[i-2])+dp[i-2]);//cout<<dp[i]<<" "<<abs(a[i]-a[i-1])<<dp[i-1]<<abs(a[i]-a[i-2])<<dp[i-2]<<endl;

}

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

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

int v[1000010];

int memo[1000010];

int c;



int f(int i)

{



    if(memo[i]!=-1)

        return memo[i];



    if(i == 0)

        return memo[0] = 0;



    if(i == 1)

        return memo[1] = abs(v[0] - v[1]);



    return memo[i] = min(abs(v[i] - v[i-1]) + f(i-1), abs(v[i] - v[i-2]) + f(i-2));

}



int main()

{



    memset(memo, -1, sizeof(memo));

    cin>>c;



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

    {

        cin>>v[i];

    }



    cout<<f(c-1)<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.204426703142244x):


#include<bits/stdc++.h>

#define ll long long int

#define s(l) scanf("%lld",&l)

#define ss(lc) scanf("%s",lc)



#define ppp push_back

#define p(x) printf("%lld\n",x)

#define sr string

#define vr vector

/*int ff(int w)

{

    if(w==1){return 1;}

    else if(w==0){return 1;}

    else{return w*ff(w-1);}

}*/

/*int pr(int ww)

{

    int xs=ww/2;

    int sx=1;

    for(int i=2;i<=xs;i++)

    {

        if(ww%i==0){sx=0;break;}



    }

    return sx;

}*/

ll a[200005],b[200005];

using namespace std;

int main()

{

    vr<ll> mn;

    memset(a,0,sizeof(a));

    memset(b,0,sizeof(b));

  ll x=0,y,z=0,zz,i,j,k=0,l,m,n,o,p,q,r,ss;

  s(n);

vr<ll>dp(n,0);

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

{

    s(a[i]);

}

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

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

{

    dp[i]=min(abs(a[i]-a[i-1])+dp[i-1],abs(a[i]-a[i-2])+dp[i-2]);//cout<<dp[i]<<" "<<abs(a[i]-a[i-1])<<dp[i-1]<<abs(a[i]-a[i-2])<<dp[i-2]<<endl;

}

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

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define ll long long

#define endl "\n"

using namespace std;

inline int D()

{

    int t;

    scanf("%d",&t);

    return t;

}

inline ll llD()

{

    ll t;

    scanf("%lld",&t);

    return t;

}



const int Max=1e6+9;

const int OO=-1e6;

int st[4]= {1,0,0,-1};

int nt[4]= {0,1,-1,0};

int n,m;

bool valid(int x,int y)

{

    return x>-1 && y>-1 && x<n && y<m;

}

ll mem[Max];

void init()

{

    memset(mem,-1,sizeof mem);

}

//////////////////////////////////////////////////////////



ll a[Max];



ll solve (int i )

{

    if(i==n-1)return 0;

    if(i>=n)

        return Max;



    ll &ret=mem[i];

    if(~ret)return ret;



    ll take1 =solve(i+1)+abs(a[i]-a[i+1]);

    ll take2 =solve(i+2)+abs(a[i]-a[i+2]);



     return mem[i]=min(take2,take1);

}



void Fast()

{

    std::ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    cerr.tie(0);

}

int main()

{

   init();

   n =D();

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

   {

       a[i]=llD();

   }

   printf("%lld",solve(0));



}


// Faster C++ solution for the same problem (measured speedup: 2.2193867049580054x):


#include<bits/stdc++.h>

#define ll long long int

#define s(l) scanf("%lld",&l)

#define ss(lc) scanf("%s",lc)



#define ppp push_back

#define p(x) printf("%lld\n",x)

#define sr string

#define vr vector

/*int ff(int w)

{

    if(w==1){return 1;}

    else if(w==0){return 1;}

    else{return w*ff(w-1);}

}*/

/*int pr(int ww)

{

    int xs=ww/2;

    int sx=1;

    for(int i=2;i<=xs;i++)

    {

        if(ww%i==0){sx=0;break;}



    }

    return sx;

}*/

ll a[200005],b[200005];

using namespace std;

int main()

{

    vr<ll> mn;

    memset(a,0,sizeof(a));

    memset(b,0,sizeof(b));

  ll x=0,y,z=0,zz,i,j,k=0,l,m,n,o,p,q,r,ss;

  s(n);

vr<ll>dp(n,0);

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

{

    s(a[i]);

}

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

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

{

    dp[i]=min(abs(a[i]-a[i-1])+dp[i-1],abs(a[i]-a[i-2])+dp[i-2]);//cout<<dp[i]<<" "<<abs(a[i]-a[i-1])<<dp[i-1]<<abs(a[i]-a[i-2])<<dp[i-2]<<endl;

}

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

}