stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.8045685481558125x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

}
// 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: 1.8113818577823777x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

}
// 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: 3.349526821499604x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

}
// 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: 1.3836968883367324x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

}
// 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: 1.3863375376509772x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

}
// 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: 9.265899979023402x):
#include<bits/stdc++.h>

#define ll long long

#define Test int t;cin>>t;while(t--)

#define PI acos(-1)

#define endl "\n"

#define fx(x) fixed<<setprecision(x)

#define sz(s) s.size()

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

#define allr(v) (v).rbegin(),(v).rend()

const double eps=1e-9;

const int mod=1e9+7;

int dx[]={-1,-1,-1,0,0,1,1,1};

int dy[]={-1,0,1,-1,1,-1,0,1};

ll gcd(ll x,ll y) {return(!y)?x:gcd(y,x%y);}

ll lcm(ll x,ll y) {return((x/gcd(x,y))*y);}

using namespace std;

void fast(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

}

const int N=(int)1e5+10;

int dp[3][N];

int a[N],n;

int solve(int p,int idx){

    if(idx>=n)return 1000000;

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

    int &ret=dp[p][idx];

    if(ret!=-1)return ret;

    if(p==1)ret=abs(a[idx+1]-a[idx]),idx++;

    else ret=abs(a[idx+2]-a[idx]),idx+=2;

    ret+=min(solve(1,idx),solve(2,idx));

    return ret;

}

int main()

{

    fast();

    cin>>n;

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

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

            dp[i][j]=-1;

        }

    }

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

        cin>>a[i];

    }

    cout<<min(solve(1,0),solve(2,0));

    return 0;

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

using namespace std;

const int mx = 100005;

int dp[mx];

int A[mx];

void memo(int i,int n)

{

	if(i>n) return;

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

	memo(i+1,n);

}

int main()

{

     int n;cin>>n;

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

     	cin>>A[i];

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

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

     memo(3,n);

     cout<<dp[n]<<endl;

     return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3697336376797706x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



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



// minCost(i) = min( abs(a[i]-a[i-1]) + minCost(i-1), abs(a[i]-a[i-2]) + minCost(i-2))

// minCost(2) = abs(a[1]-a[0])

// minCost(3) = min(abs(a[2]-a[0]), abs(a[1]-a[2]) + abs(a[0]-a[1])



class BeingZero{

    public:

        ll minCost(vector<int> a){

            ll n = a.size();

            int dp[n];

            dp[0] = 0;

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

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

            //dp[3] = min(abs(a[2]-a[0]), abs(a[1]-a[0]) + abs(a[2]-a[1]));

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

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

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

            //     cout<<i<<" : "<<dp[i]<<"\n";

            return dp[n-1];

        }

};





int main() {

    fastio;

    BeingZero bz;

    int n;

    while(cin>>n){

        if(n==-1) break;

        vector<int> v(n);

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

            cin>>v[i];

        cout<<bz.minCost(v)<<"\n";

    }

    return 0;

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

using namespace std;



#define INF 1e18;

#define MAX_N 100005

long long N, h[MAX_N], memo[MAX_N];



long long dp(int pos) {

  if (pos < 1) {

  	return INF;

  }

  if (pos == 1) {

  	return 0;

  }

  if (pos == 2) {

  	return abs(h[2] - h[1]);

  }

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

  	return memo[pos];

  }

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

}



int main() {

  cin >> N;

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

  	cin >> h[i];

  }

  /*

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

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

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

  }

  */

  /*

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

  	cout << i << ' ' << memo[i] << '\n';

  }

  */

  // cout << memo[N] << '\n';

  memset(memo, -1, sizeof memo);

  cout << dp(N) << '\n';

}

// Faster C++ solution for the same problem (measured speedup: 1.760755375917387x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



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



// minCost(i) = min( abs(a[i]-a[i-1]) + minCost(i-1), abs(a[i]-a[i-2]) + minCost(i-2))

// minCost(2) = abs(a[1]-a[0])

// minCost(3) = min(abs(a[2]-a[0]), abs(a[1]-a[2]) + abs(a[0]-a[1])



class BeingZero{

    public:

        ll minCost(vector<int> a){

            ll n = a.size();

            int dp[n];

            dp[0] = 0;

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

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

            //dp[3] = min(abs(a[2]-a[0]), abs(a[1]-a[0]) + abs(a[2]-a[1]));

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

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

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

            //     cout<<i<<" : "<<dp[i]<<"\n";

            return dp[n-1];

        }

};





int main() {

    fastio;

    BeingZero bz;

    int n;

    while(cin>>n){

        if(n==-1) break;

        vector<int> v(n);

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

            cin>>v[i];

        cout<<bz.minCost(v)<<"\n";

    }

    return 0;

}
// 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: 4.96148196099273x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



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



// minCost(i) = min( abs(a[i]-a[i-1]) + minCost(i-1), abs(a[i]-a[i-2]) + minCost(i-2))

// minCost(2) = abs(a[1]-a[0])

// minCost(3) = min(abs(a[2]-a[0]), abs(a[1]-a[2]) + abs(a[0]-a[1])



class BeingZero{

    public:

        ll minCost(vector<int> a){

            ll n = a.size();

            int dp[n];

            dp[0] = 0;

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

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

            //dp[3] = min(abs(a[2]-a[0]), abs(a[1]-a[0]) + abs(a[2]-a[1]));

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

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

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

            //     cout<<i<<" : "<<dp[i]<<"\n";

            return dp[n-1];

        }

};





int main() {

    fastio;

    BeingZero bz;

    int n;

    while(cin>>n){

        if(n==-1) break;

        vector<int> v(n);

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

            cin>>v[i];

        cout<<bz.minCost(v)<<"\n";

    }

    return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f3f3f3f

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define pb push_back

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

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

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

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int mod = 1e9;

const int N = 1e5+10;

/*=======================================================================*/



int n, h[N], sum;

int f[N];



int main() {

   scanf("%d", &n);

   for1(i, n) {

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

   }

   ms(f, inf);

   f[n] = 0;

   for(int i = n; i > 1; i--) {

      if(i == 2) {

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

      } else {

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

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

      }

      // if(i == 2) {

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

      //    i--;

      // } else {

      //    if(abs(h[i] - h[i-1]) < abs(h[i] - h[i-2])) {

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

      //       i -= 1;

      //    } else {

      //       sum += abs(h[i] - h[i-2]);

      //       i -= 2;

      //    }

      // }

   }

   cout << f[1] << endl;

   return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3607494778215952x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



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



// minCost(i) = min( abs(a[i]-a[i-1]) + minCost(i-1), abs(a[i]-a[i-2]) + minCost(i-2))

// minCost(2) = abs(a[1]-a[0])

// minCost(3) = min(abs(a[2]-a[0]), abs(a[1]-a[2]) + abs(a[0]-a[1])



class BeingZero{

    public:

        ll minCost(vector<int> a){

            ll n = a.size();

            int dp[n];

            dp[0] = 0;

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

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

            //dp[3] = min(abs(a[2]-a[0]), abs(a[1]-a[0]) + abs(a[2]-a[1]));

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

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

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

            //     cout<<i<<" : "<<dp[i]<<"\n";

            return dp[n-1];

        }

};





int main() {

    fastio;

    BeingZero bz;

    int n;

    while(cin>>n){

        if(n==-1) break;

        vector<int> v(n);

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

            cin>>v[i];

        cout<<bz.minCost(v)<<"\n";

    }

    return 0;

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

using namespace std;



const int nax = 1e5+3;

int n;

int dp[nax];

int a[nax];

int f(int i){

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

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

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

	if(i!=n-2) ans = min(ans, abs(a[i]-a[i+2])+f(i+2));

	return dp[i]=ans;

}

int main(){

	cin>>n;

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

	memset(dp, -1, sizeof dp);

	cout<<f(0)<<endl;

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3696902007055944x):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;



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



// minCost(i) = min( abs(a[i]-a[i-1]) + minCost(i-1), abs(a[i]-a[i-2]) + minCost(i-2))

// minCost(2) = abs(a[1]-a[0])

// minCost(3) = min(abs(a[2]-a[0]), abs(a[1]-a[2]) + abs(a[0]-a[1])



class BeingZero{

    public:

        ll minCost(vector<int> a){

            ll n = a.size();

            int dp[n];

            dp[0] = 0;

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

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

            //dp[3] = min(abs(a[2]-a[0]), abs(a[1]-a[0]) + abs(a[2]-a[1]));

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

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

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

            //     cout<<i<<" : "<<dp[i]<<"\n";

            return dp[n-1];

        }

};





int main() {

    fastio;

    BeingZero bz;

    int n;

    while(cin>>n){

        if(n==-1) break;

        vector<int> v(n);

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

            cin>>v[i];

        cout<<bz.minCost(v)<<"\n";

    }

    return 0;

}