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

#include <vector>

using namespace std;

typedef long long ll;



template<class T> inline bool chmin(T& a, T b){

    if(a>b){

        a=b;

        return true;

    }

    else

    {

        return false;

    }

}



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

{

    int n;

    const ll INF = 1ll<<60;

    vector<ll> dp(100000,INF);

    cin >> n;

    vector<int> h(n);

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

    dp[0]=0;

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

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

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

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

    }

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

    



    return 0;

}


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

#include <vector>

using namespace std;



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

{

    int n;cin>>n;

    vector<int> h(n);

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

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

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

    if (n==2){

        cout << dp[1]<< endl;

        return 0;

    }

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

    if (n==3){

        cout << dp[2]<< endl;

        return 0;

    }

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

        int d1 =dp[i-1]+abs(h[i]-h[i-1]);

        int d2 =dp[i-2]+abs(h[i]-h[i-2]);

        dp[i]=min(d1,d2) ;

    }

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



    return 0;

}


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



using namespace std;



typedef long long ll;

typedef vector <int> vi;

typedef vector <ll> vl;

typedef pair <ll, ll> dl;

typedef pair <int, int> ii;

typedef vector <ii> vii;

typedef vector <vi> graph;



int n;

ll arr[100010];

ll dp[100010];



ll piedra(int id) {

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

	if (id >= n) return 1000000000;

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

	ll temp1, temp2;

	temp1 = abs(arr[id] - arr[id + 1]) + piedra(id + 1);

	temp2 = abs(arr[id] - arr[id + 2]) + piedra(id + 2);

	return dp[id] = min(temp1, temp2);

}



int main() {

    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);



	cin >> n;

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

	{

		cin >> arr[i];

	}

	for (int i = 0; i < 100010; ++i)

	{

		dp[i] = -1;

	}

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

    return 0;

}


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

#define tam 100010

#define inf 1000000000



using namespace std;



typedef long long ll;

typedef vector <int> vi;

typedef vector <ll> vl;

typedef pair <ll, ll> dl;

typedef pair <int, int> ii;

typedef vector <ii> vii;

typedef vector <vii> graph;



bool comp(int a, int b) {

	return a > b;

}



int main() {

	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);



	int n;

	cin >> n;

	int arr[n];

	vi dp (n, inf);

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

	{

		cin >> arr[i];

	}

	dp[0] = 0;

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

	{

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

		{

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

		}

	}

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

	return 0;

}


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



using namespace std;

using ll = long long;



using Field = vector<vector<int>>;

using Graph = vector<vector<int>>;

using VI = vector<int>;

using VC = vector<char>;

using PI = pair<int, int>;



#define REP(i, n) for (int i = 0; i < (n); i++)

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



const long long INF = 1LL<<60;

const int mod = 1000000007;



template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }



int N;

vector<int> H(100010);

vector<int> H1(100010);

vector<int> H2(100010);

int MAX_N = 100000;



vector<ll> dp(100010, INF);



int main() {

    cin.tie(0);

    ios_base::sync_with_stdio(false);

    cin >> N;

    REP(i, N) {

        cin >> H.at(i);

    }

    for (int i = 0; i < N - 1; i++) {

        H1.at(i) = abs(H.at(i) - H.at(i + 1));

    }

    for (int i = 0; i < N - 2; i++) {

        H2.at(i) = abs(H.at(i) - H.at(i + 2));

    }



    dp.at(0) = 0;



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

        chmin(dp.at(i + 1), dp.at(i) + H1.at(i));

        chmin(dp.at(i + 2), dp.at(i) + H2.at(i));

    }



    cout << dp.at(N-1) << endl; 



    return 0;

}

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



using namespace std;

using ll = long long;



using Field = vector<vector<int>>;

using Graph = vector<vector<int>>;

using VI = vector<int>;

using VC = vector<char>;

using PI = pair<int, int>;



#define REP(i, n) for (int i = 0; i < (n); i++)

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



const long long INF = 1LL<<60;

const int mod = 1000000007;



template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }



int N;

vector<int> h;

vector<ll> dp;



int main() {

    cin.tie(0);

    ios_base::sync_with_stdio(false);



    cin >> N;

    h.resize(N + 10, 0);

    dp.resize(N + 10, INF);



    REP(i, N) cin >> h.at(i);



    dp[0] = 0;



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

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

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

    }



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

    

    return 0;

}

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

#include <bits/stdc++.h>

using namespace std;

int static dp[100001];

int frog(vector<int>&a,int n,int current)

{

    if(current+1==n)

        return 0;

    if(dp[current]!=-1)

        return dp[current];

    int A=INT_MAX,B=INT_MAX;

    if(current+1<n)

        A=abs(a[current]-a[current+1])+frog(a,n,current+1);

    if(current+2<n)

        B=abs(a[current]-a[current+2])+frog(a,n,current+2);

    dp[current]=min(A,B);

    return dp[current];

}



int main() {

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

	int n;

	cin>>n;

	vector<int>height;

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

	{

	    int x;

	    cin>>x;

	    height.push_back(x);

	}

	cout<<frog(height,n,0);

	return 0;

}

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

#include <bits/stdc++.h>

using namespace std;

int main() {

	int n;

	cin>>n;

	vector<int>height;

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

	{

	    int x;

	    cin>>x;

	    height.push_back(x);

	}

	vector<int>dp(n,INT_MAX);

	dp[0]=0;

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

	{

	    for(int j=i+1;j<=i+2;j++)

	    {

	        if(j<n)

	        {

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

	        }

	    }

	}

	cout<<dp[n-1];

	return 0;

}

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

#include<string>

#include<cstdio>

#include<algorithm>

#include<cstdlib>

#include<cmath>

#include<vector>

#include<utility>

#include<queue>

#include<stack>

#include<set>

#include<map>

#include<unordered_set>

#include<bitset>

using namespace std;

//#define MODE 1

#ifdef MODE

#define DEB(X) cout<< #X <<": "<<X<<" ";

#define ARDEB(i,X) cout<< #X <<"["<<i<<"]: "<<X[i]<<" ";

#define END cout<<endl;

#else

#define DEB(X) {}

#define ARDEB(i,X) {}

#define END {}

#endif

typedef long long int ll;

#define int long long

typedef pair<int,int> P;

struct edge{int to,cost;};

#define REP(i,n) for(int i=0;i<(n);i++)

const int INF=100000000000000000;



int N,A[100010];

int dp[100010];



int DP(int a){

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

	int res=INF,res2=INF;

	if(a==N-1){

		return 0;

	}

	res=DP(a+1)+abs(A[a]-A[a+1]);

	if(a<N-2){

		res2=DP(a+2)+abs(A[a+2]-A[a]);

	}

	return dp[a]=min(res,res2);

}



signed main(){

	cin>>N;

	REP(i,N){

		cin>>A[i];

	}

	REP(i,100010){

		dp[i]=-1;

	}

	cout<<DP(0)<<endl;

	return 0;

}


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

#include<string>

#include<cstdio>

#include<algorithm>

#include<cstdlib>

#include<cmath>

#include<vector>

#include<utility>

#include<queue>

#include<stack>

#include<set>

#include<map>

#include<unordered_set>

#include<unordered_map>

#include<bitset>

#include<iomanip>

#include<list>

#include<deque>

using namespace std;

//#define MODE 1

#ifdef MODE

#define DEB(X) cout<< #X <<": "<<X<<" ";

#define END cout<<endl;

#else

#define DEB(X) {}

#define END {}

#endif

typedef long long ll;

#define int ll

#define uset unordered_set

#define umap unordered_map

typedef std::pair<int,int> P;

struct edge{int to,cost;};

const int INF=100000000000000000;

const int INF2=9223372036854775807;

const int MOD=1000000007;

#define REP(i,n) for(int i=0;i<(int)(n);i++)

#define min(X,Y) (((int)(X)<(int)(Y))?(X):(Y))

#define max(X,Y) (((int)(X)>(int)(Y))?(X):(Y))

#define MAX(X,Y) (*max_element(X,Y))

#define MIN(X,Y) (*min_element(X,Y))

#define NP(X,Y) next_permutation(X,Y)

#define setp(X,Y) cout<<fixed<<setprecision(Y)<<X;

int ceil2(int a,int b){if(a%b==0){return a/b;}else{return a/b+1;}}

int pow2(int a,int b){int r=1;for(int i=1;i<=b;i++){r*=a;}return r;}

int Log2(int a){int t=0;while(1){if(a==0||a==1){break;}a/=2;t++;}return t;}



int N;

int H[100010];

int dp[100010];



signed main(){

	cin>>N;

	REP(i,N){

		cin>>H[i];

		dp[i]=INF;

	}

	dp[0]=0;

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

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

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

	}

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

	return 0;

}


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



#define F first

#define S second

#define PB push_back



using namespace std;



typedef long long ll;



int n;

ll h[100005];

ll memo[100005];



ll dp(int i){

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

	if(i >= n)	return 10000000000;

	if(memo[i] != -1){

		return memo[i];

	}

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

}



int main(){



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

	

	ios_base::sync_with_stdio(false);

	cin.tie(NULL);

	cout.tie(NULL);



	cin >> n;

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

		cin >> h[i];

	}



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



	return 0;

}

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



#define F first

#define S second

#define PB push_back



using namespace std;



typedef long long ll;



int main(){

	

	ios_base::sync_with_stdio(false);

	cin.tie(NULL);

	cout.tie(NULL);



	int n;

	cin >> n;



	int h[n];

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

		cin >> h[i];

	}



	int cost[n];

	cost[0] = 0;

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



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

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

	}



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



	return 0;

}

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

#define rep(i, n) for(int i = 0; i < (int)n; i++)

using namespace std;

typedef long long ll;



template<class T> inline bool chmax(T& a, T b){

    if (a < b) {

        a = b;

        return true;

    }

    return false;

}



template<class T> inline bool chmin(T& a, T b) {

    if (a > b) {

        a = b;

        return true;

    }

    return false;

}



const ll INF = 1LL << 60;



ll h[100010];



ll dp[100010];



ll rec(int i){

    if(dp[i] < INF) return dp[i];

    

    if(i == 0) return 0;

    

    ll res = INF;

    

    chmin(res, rec(i - 1) + abs(h[i] - h[i - 1]));

    if(i > 1) chmin(res, rec(i - 2) + abs(h[i] - h[i - 2]));

    

    return dp[i] = res;

}





int main(){

    

    cin.tie(0);

    ios::sync_with_stdio(0);

    

    int n;

    cin >> n;

    

    rep(i,n) cin >> h[i];

    

    rep(i,100010) dp[i] = INF;

    

    cout << rec(n - 1) << endl;



}


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

using namespace std;

#define rep(i, n) for(int i = 0; i < (int)n; i++)

typedef long long ll;



int dp[100010];



int main(){

    cin.tie(0);

    ios::sync_with_stdio(false);



    int n;

    cin >> n;

    

    vector<int> h(n);

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

    

    

    

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

        if(i == 0) continue;

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

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

    }

    

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

}


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

using namespace std;

#define N (int)2e5

int n,memo[N],c[N];



int dp(int curr) {

    if (curr == n) return 0;

    if (curr > n) return 1e9;

    int &sol = memo[curr];

    if (sol != -1) return sol;

    sol = min(

        dp(curr + 1) + abs(c[curr] - c[curr + 1]) ,

        dp(curr + 2) + abs(c[curr] - c[curr + 2])

    );

    return sol;

}



int main() {

    scanf("%d" , &n);

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

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

    memset (memo , -1 , sizeof memo);

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

}


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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

// Fastest known solution for this problem:
// 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 + 5;//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;

int h[maxn], dp[maxn];



int main()

{

	n = read();

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

	memset(dp, 0x3f, sizeof dp);

	dp[1] = 0;

	rep(i, 1, n){

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

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

	}

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

	return 0;

}

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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

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


/**

    And in the end, the love you take is equal to the love you make.



        I LOVE YOU LAURA.

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rope>



#define _CRT_SECURE_NO_DEPRECATE

#define _CRT_DISABLE_PERFCRIT_LOCKS



#define findOrder find_by_order

#define orderKey order_of_key



#define isOn(mask, i) (mask & (1 << i))

#define turn(mask, i) (mask |= (1 << i))

#define switch(mask, i) (mask &= ~(1 << i))

#define change(mask, i) (mask ^= (1 << i))

#define lowbit(mask) (mask & (-mask))

#define turnAll(mask, n) (mask = (1 << n) - 1)

#define switchLastBit(mask) (mask & (mask - 1))

#define turnLastZero(mask) (mask | (mask + 1))

#define modulo(x, mod) ((x) & (mod - 1))

#define log2(x) (31 - __builtin_clz(x))



#define mkp make_pair

#define pb push_back

#define pf push_front

#define eb emplace_back

#define ef emplace_front

#define ins insert

#define rsz resize

#define f first

#define s second

#define lb lower_bound

#define ub upper_bound



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

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

#define rall(x) rbegin(x), rend(x)



#define debug(x) (cerr << x << " ")



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef double db;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;



typedef pair<int, int> ii;

typedef pair<ll,ll> pll;

typedef pair<ld,ld> pdb;

typedef pair<int, ii> tri;

typedef complex<ld> cd;



typedef vector<int> vi;

typedef vector<ii> vii;

typedef vector<ld> vdb;

typedef vector<cd> vcd;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<vll> matrix;



typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_tree;



const ll mod = 1e9+7;

const int maxn = 2e5+2;

const int pi = 2 * acos(0.0);

const db eps = 1e-9;



ll arr[maxn], dp[maxn];



int main()

{

    ios_base::sync_with_stdio(0);

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



    int n; cin >> n;



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

        cin >> arr[i];

    }



    memset(dp, 63, sizeof dp);



    dp[1] = 0;

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

        if (i <= 2){

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

        }

        else{

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

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

        }

    }



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



    return 0;

}


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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

    printf("%d",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 int

#define dbl double

#define pi acos(-1)

#define sc1(a) scanf("%lld",&a)

#define sc2(a,b) scanf("%lld%lld",&a,&b)

#define sc3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)

#define pb push_back



ll ara[1000005];



map<ll,ll>vis;

map<ll,ll>vis1;

map<ll,ll>pos;

map<ll,ll>high;

map<ll,ll>low;

map<string,ll>mp;

vector<ll>vec;

vector<ll>v;

stack<ll>q;

string str1[100];

string str2[100];

priority_queue<ll>pq;

ll dp[1000005],t;

ll mn=1000000;

ll call(ll id)

{

    if(id==t )return 0;

    if(dp[id]!=-1)

        return dp[id];

    ll p=call(id+1)+abs(ara[id+1]-ara[id]);

    ll q=100000000;

    if(id+2!=t+1)

        q=call(id+2)+abs(ara[id+2]-ara[id]);

    dp[id]=min(p,q);

    return dp[id];



}

int main()

{



    // ll t;

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

    sc1(t);

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

    {

        sc1(ara[i]);

    }

    cout<<call(1)<<endl;

    return 0;

}


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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

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



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

#define ford(i, n) for (int i = int(n)-1; i >= 0; --i)

#define pb push_back

#define eb emplace_back

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

#define F first

#define S second

#define endl '\n'

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

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

int dp[N];

int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int n;

	cin>>n;

	vector<int> a(n);

	for(auto& it:a) cin>>it;

	ms(dp,INF);

	dp[0]=0;

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

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

		if(i>1) dp[i]=min(dp[i],dp[i-2]+abs(a[i]-a[i-2]));

	}

	cout<<dp[n-1];

    return 0;

}




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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

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

using namespace std;



int a[1000001];

int dp[1000001];

int it;



int fr(int p){

	if (p==it-1){

      return 0;

    }

	if(dp[p]!=-1){

      return dp[p];

    }

	int re=abs(a[p]-a[p+1])+fr(p+1);

	if(p!=it-2){

      re=min(re,abs(a[p]-a[p+2])+fr(p+2));

    }

	return dp[p]=re;

}

int main(){

	cin>>it;

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

    cin>>a[i];

    }



	memset(dp,-1,sizeof dp);

	cout<<fr(0)<<endl;

	return 0;

}


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

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

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

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

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

// Fastest known solution for this problem: