stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define Joe ios::sync_with_stdio(0); cin.tie(NULL); ios_base::sync_with_stdio(0); cout.tie(NULL);

#define INF 1e9+7

const long long N =1e5+5;

using namespace std;

int main(){

	Joe

	int n; cin>>n;

	int arr[n+5];

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

		cin>>arr[i];

	}

	int dp[n+5];

	dp[0]=0;

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

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

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

	}

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

	return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <bits/stdc++.h>

#define mod 998244353

typedef long long ll;



using namespace std;



void Fastio(){

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

    if (exp == 0)

        return 1;

    ll ans = fp(base, exp / 2);

    ans = (ans * ans) % mod;

    if (exp % 2 != 0)

        ans = (ans * (base % mod)) % mod;

    return ans;

}



void calcFacAndInv(ll n)

{

    fact[0] = inv[0] = 1;

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

    {

        fact[i] = (i * fact[i - 1]) ;

        inv[i] = fp(fact[i], mod - 2);

    }

}



ll ncr(ll n, ll r)

{

    return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

    return (fact[n] * inv[n - r]) % mod;

}



long double  NCR(int n, int m){

    long double res = 1;

    m = min(m, n - m);

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

        res = res * ( n - i) / (i + 1);

 }

    return res;

}

ll tol(string a){

    ll res=0;

    while(a[0]=='0'){

        a.erase(a.begin());

    }

    if(a=="")a="0";

    for(int i=0; i<a.size(); i++){

        res*=10; res+=a[i]-'0';

    }

    return res;

}

int check(string a, string b){

    for(int i=0; i<a.size(); i++){

        if(a[i]>b[i])return 1;

        if(b[i]>a[i])return -1;

    }

    return 0;

}

class data{

public:

    ll v,d,p;

    data(){}

    data(ll vv, ll dd, ll pp){

        v=vv;  d=dd; p=pp;

    }

};



int primeFactors(int n)

{

    ll c=0;

    while (n % 2 == 0)

    {

        c++;

        n = n/2;

    }

    for (int i = 3; i*i <= n; i = i + 2)

    {

        while (n % i == 0)

        {

            c++;

            n = n/i;

        }

    }

    if (n > 2)

        c++;

    return c;

}



int main()

{

    Fastio();

    int n; cin>>n;

    int arr[n];

    for(int i=0; i<n; i++){cin>>arr[i];}

    ll dp[n]={0};

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

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

        ll ans1, ans2;

        ans1=abs(arr[i]-arr[i-1])+dp[i-1];

        ans2=abs(arr[i]-arr[i-2])+dp[i-2];

        dp[i]=min(ans1,ans2);

    }

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



    return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;

const int N = 1e4 + 5;



int main()

{

	int n;

	cin >> n;

	int height[n+1], dp[n+1];

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

		cin >> height[i];

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

	dp[0] = 0;

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

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

	{

		long long a, b;

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

		b = abs(height[i] - height[i - 2]) + dp[i - 2];

		dp[i] = min(a, b);

	}

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



	return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



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

#define _(x) {cout << #x << " = " << x << " ";}

#define sz(a) int((a).size())

#define pb push_back

#define F first

#define S second

#define MOD 1000000007



typedef long long ll;

typedef unsigned long long ull;



const double E = 1e-8;

const double PI = acos(-1);

    

using namespace std;



int N, h[100005], dp[100005]; 



int solve() {

    dp[0] = 0;

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

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

        if(i > 1)

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

    }

    return dp[N-1];

}

    

int main(int argc, char** argv) {

    

    IOBOOST

    

    cin>>N;

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

        cin>>h[i];

    cout<<solve()<<endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;

int dp[100005],h[100005];

int main()

{

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

    int n;scanf("%d",&n);

    for(int i=0;i<n;++i)scanf("%d",h+i);

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

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

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

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

    }

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

    return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5+10;

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

int main(){

	scanf("%d",&n);

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

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

	}

	f[1]=0;

	f[2]=abs(h[2]-h[1]);

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

		f[i]=min(f[i-1]+abs(h[i-1]-h[i]),f[i-2]+abs(h[i-2]-h[i]));

	}

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

} 

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define inf 10000

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+3];

    int dp[100001];



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

    {

        cin>>a[i];

        dp[i]=inf;

    }

    dp[0]=0;

    dp[1]=0;

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

    for(int i=3;i<n+1;i++)

    {

        int l1=dp[i-1]+abs(a[i-2]-a[i-1]);

        int l2=dp[i-2]+abs(a[i-3]-a[i-1]);



        dp[i]=min(l1,l2);

    }



    cout<<dp[n];





}




// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



int main()

{

    long long int  n;

    cin>>n;

    long long int  hi[n];



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

    {

        cin>>hi[i];

    }

    long long int  ans[n];

    ans[0] = 0;

    ans[1] = abs(hi[1] - hi[0]);

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

    {

        ans[i] =  min(ans[i - 2]   + abs(hi[i] - hi[i - 2]),ans[i - 1] + abs(hi[i] -hi[i - 1]));



    }

    cout<<ans[n-1];

    return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

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

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

#define clr(v, d) memset(v, d, sizeof(v))

#define fi first

#define se second

#define pp push_back

#define pb pop_back

#define pi 3.141592654

#define MOD 1000000007 

#define cons1 1e12

#define cons2 1e6

using namespace std;

typedef long long ll;

typedef vector<int> vi;

typedef pair<int, int> pa;

typedef unsigned long long int ull;

typedef double long dl;

const int N = 1e5+9;

int n;

int v[N], ans[N];

int sol(int i) {

	if (ans[i] == -1) {

		if (i == n - 1)

			ans[i] = 0;

		else if (i == n - 2)

			ans[i] = abs(v[i] - v[i + 1]);

		else 

			ans[i] = min(abs(v[i] - v[i + 1])+sol(i + 1)

				, abs(v[i] - v[i + 2]) + sol(i + 2));

		

	}

	return ans[i];

	

}

void solve() {

	clr(ans, -1);

	cin >> n;

	rep(i, 0, n)

		cin >> v[i];

	cout << sol(0) << endl;

}

int main() {

	ios::sync_with_stdio(0);

	cin.tie(NULL);

	cout.tie(NULL);

	int t = 1;

	//cin >> t;

	while (t--)

		solve();



#ifndef ONLINE_JUDGE

	- system("pause");

#endif



}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <algorithm>

using std::min;



#define MAXN 100003



int abs(int z)

{

    if (z<0)

        return -z;

    else

        return z;

}



int main(void)

{

    int N,i,j;

    int h[MAXN], p[MAXN];

    scanf("%d",&N);

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

    {

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

    }

    p[N-1]=0;

    p[N-2]=abs(h[N-2]-h[N-1]);

    for(i=N-3;i>=0;i--)

    {

        p[i]=min(p[i+1]+abs(h[i]-h[i+1]),p[i+2]+abs(h[i]-h[i+2]));

    }

    printf("%d\n",p[0]);



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



typedef long double LD;

typedef long long ll;

typedef map<int,int> mii;

typedef pair<int,int> pii;

typedef vector<pii> vpii;



#define pb              push_back

#define mp              make_pair

#define F               first

#define S               second

#define mod             1000000007

#define INF             2147483647

#define Endl            "\n"

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

#define precision(a)    cout << fixed << setprecision(a)

#define bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define reset(a,b)      memset(a,b,sizeof(a));

#define srt(a)      sort(a.begin(),a.end());



const int MAXN = 1000005;

const int SQRTN = 1003;

const int LOGN = 22;

const double PI=acos(-1);

const int sz=1e5+9;



int strtoint(string str){

    stringstream ss(str);

    int x = 0;

    ss >> x ;

    return x ;

}

string intostr(int x){

    stringstream ss;

    ss << x;

    string str = ss.str();

    return str;

}

void rem(string &s, const string &c) {

	auto pos = s.find(c);

	if (pos == string::npos) {

		cout << -1 << endl;

		exit(0);

	}

	s.erase(0, pos + 1);

}

void println (int n){

    cout<<n<<"\n";

}

void printvec(vector <int> &a){

    for (int i=0;i<a.size();i++){

        cout<<a[i]<<" ";

    }

    cout<<"\n";

}



void printgraph(vector <int> g[],int e){

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

        cout<<i<<" --> ";

        for (int j=0;j<g[i].size();j++){

            cout<<g[i][j]<<" ";

        }

        cout<<"\n";

    }

}



void printmap (map <int,int> &mymap){

    for (auto itr = mymap.begin(); itr != mymap.end(); ++itr) {

        cout << '\t' << itr->first

             << '\t' << itr->second << '\n';

    }

    cout << "\n";

}



int n,x,arr[sz],dp[sz];





int rec (int i){

    if (i==n) return 0;

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

    int left = abs(arr[i]-arr[i+1])+rec(i+1);

    int right = INF;

    if (i+2<=n){

        right = abs(arr[i]-arr[i+2])+rec (i+2);

    }

    return dp[i]= min(left,right);

}





void solve(){



    cin>>n;

    reset (dp,-1);

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





    cout << rec (1) << Endl;



    //printvec(v);

}





int main(){

    //FasterIO;



    solve();

    return 0;

}


// Optimized solution to the same problem (faster):
#include<iostream>

#include<cmath>

using namespace std;

const int maxn = 1e6 + 10;

int height[maxn];

long long dp[maxn];



int main(){

    int n;

    scanf("%d", &n);

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

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

    dp[1] = 0;

    dp[2] = abs(height[2] - height[1]);

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

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

    printf("%lld", dp[n]);

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



const int N = 1e5; // Maximum stones as per problem



int n, stones[N];



int memo[N]; // Memoization Table



int jumps(int ind) {

    if (ind == (n - 1)) return 0; // If we are last stone, then the answer will be 0.

    if (ind >= n) return INT_MAX; // Checking the boundary condition





    // If you are done with the whole function, then only read

    // Since we stored the minimum cost for a stone, now we will use the stored cost

    // We are initializing the table to -1, if the value stored is not -1, then we have calculated the minimum cost

    // So just return that cost

    if (memo[ind] != -1) return memo[ind]; // If reading for first time, then ignore this line



    // Calculate the cost for jumping to next stone (ind + 1) and recursively doing it for next stone

    int oneJump = abs(stones[ind] - stones[ind + 1]) + jumps(ind + 1);



    // Initializing the twoJumps because there can be a case in which there can be only one stone.

    int twoJumps = INT_MAX;



    // Checking if there are more than 2 stones

    if (ind + 2 < n) {

        // Calculate the cost for jumping to next to next stone (ind + 2) and recursively doing it for next to next stone

        twoJumps = abs(stones[ind] - stones[ind + 2]) + jumps(ind + 2);

    }



    // Storing the minimum cost in the memoization table and returning the minimum cost

    // After this,  go to the ignored line.

    return memo[ind] = min(oneJump, twoJumps);

}





int main() {

    // Initializing the memoization table to -1

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



    cin >> n;

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

        cin >> stones[i];

    }



    // We will start from first stone

    cout << jumps(0) << endl;



    return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cmath>

using namespace std;

const int maxn = 1e6 + 10;

int height[maxn];

long long dp[maxn];



int main(){

    int n;

    scanf("%d", &n);

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

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

    dp[1] = 0;

    dp[2] = abs(height[2] - height[1]);

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

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

    printf("%lld", dp[n]);

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define INF 1e18

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

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

typedef vector<int> vi;

typedef vector<ll> vll;

typedef pair<int,int> pi;



#define M_PI 3.14159265358979323846



int a[100005];

const int oo = 1e9+10;

const int mod = 7901;



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



int dp[100005];

int n;

int memo[100005];



int dpF(int i){

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

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

	

	int ans = abs(a[i] - a[i+1])+dpF(i+1);

	if(i+2 <= n-1){

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

	}



	return memo[i] = ans;

}



void solve(){

	cin>>n;

	for(int i = 0 ; i < n ; ++i)cin>>a[i];

	

	memset(memo, -1 , sizeof memo);

	cout << dpF(0) << '\n';

	return;

	dp[0] = 0;

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

	for(int 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[n-1] << '\n';

}

int main(){   

    ios::sync_with_stdio(0);cin.tie(0);

	int t=1;

	while(t--){

		solve();

	}

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cmath>

using namespace std;

const int maxn = 1e6 + 10;

int height[maxn];

long long dp[maxn];



int main(){

    int n;

    scanf("%d", &n);

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

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

    dp[1] = 0;

    dp[2] = abs(height[2] - height[1]);

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

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

    printf("%lld", dp[n]);

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n;

int h[MXN];

int dp[MXN];



int Dp(int idx) {

    if(idx == n - 1) return 0; // caso base

    if(dp[idx] != -1) return dp[idx]; // caso dp

    int ans = Dp(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 < n) ans = min(ans, Dp(idx + 2) + abs(h[idx] - h[idx + 2])); 

    return dp[idx] = ans; // guardar

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

        cin >> h[i];

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0) << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cmath>

using namespace std;

const int maxn = 1e6 + 10;

int height[maxn];

long long dp[maxn];



int main(){

    int n;

    scanf("%d", &n);

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

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

    dp[1] = 0;

    dp[2] = abs(height[2] - height[1]);

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

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

    printf("%lld", dp[n]);

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



const int inf=0x3f;

int h[100005],dp[100005];



signed main()

{

	int n;

	while(cin>>n){

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

		memset(dp,inf,sizeof dp);

		

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

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

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

		}

		cout<<dp[n]<<endl;

	}

	return 0;

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<cmath>

using namespace std;

const int maxn = 1e6 + 10;

int height[maxn];

long long dp[maxn];



int main(){

    int n;

    scanf("%d", &n);

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

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

    dp[1] = 0;

    dp[2] = abs(height[2] - height[1]);

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

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

    printf("%lld", dp[n]);

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

vector<int> dp(100001,-1);



int back(vector<int> &N, int index){

	if(index+2==N.size()){

		return abs(N[index]-N[index+1]);

	}

	if(index+1==N.size()){

		return 0;

	}

	if(dp[N.size()-index]!=-1)return dp[N.size()-index];

	int a = abs(N[index]-N[index+1]) + back(N,index+1);

	int b = abs(N[index]-N[index+2]) + back(N,index+2);

	

	dp[N.size()-index] = min(a,b);

	return dp[N.size()-index];

}



int main(){

//	10 30 40 20

	int n;

	cin>>n;

	vector<int> N(n);

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

		cin>>N[i];

	}

	dp[0]=0;

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



	cout<<back(N,0);	

}

// Optimized solution to the same problem (faster):
#include <iostream>



using namespace std;



int N, ns[100000+9];

int dp[100000+9];

int main(){

    scanf("%d",&N);

    for(int i=0;i<N;i++)scanf("%d",&ns[i]);

    dp[0]=0, dp[1]=abs(ns[0]-ns[1]);

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

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

    }

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

    return 0;

}


// Fastest known solution for this problem: