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

#include <bitset>

using namespace std;

#define ll long long

#define mod 1000000009



// يلعن ابو الي يعقدها على نفسه تاني



int main()

{

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



    ll n; cin >> n;

    ll arr[100005];

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

        cin >> arr[i];

    }

    if(n == 2){

        cout << abs(arr[1]-arr[0]);

    }else{

        ll sol[100005] = {};

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

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

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

        }









        cout << sol[n-1];

    }















    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.5883698266967596x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

 

using namespace std;

 

#define sf(n) scanf("%d", &n)

#define pd(n) printf("%.12lf", n)

#define pf(n) printf("%d", n)

#define p() printf("\n")

#define ps() printf(" ")

#define ll long long

#define ull unsigned long long

#define MAX 1e9

const int mod = 1000 * 1000 * 1000 + 7;

#define ex 1e18

#define dbl 2e09

#define qu(q) queue<int> q

#define pqu(q) priority_queue<int> q

#define len(s) s.length()

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

#define pb(x) push_back(x)

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

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

#define F first

#define S second

#define clr(x) x.clear()

#define LOG(x) std::cout << x << std::endl;

 

const int INF = 0x3f3f3f3f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const double PI = 3.14159265359;

 

int dx[6] = {-1,-1,0,0,1,1};

int dy[6] = {-1,0,-1,1,0,1};



int a[100005];

int dp[100005];

int n;

int solve(int idx){

    if(idx==(n-1))

        return 0;

    if(idx>=n)

        return INF;

    int &ret = dp[idx];

    if(~ret)

        return ret;

    ret = INF;

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

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

    return ret;

}







int main(){



    cin>>n;

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

        cin>>a[i];

    }

    memset(dp,-1,sizeof dp);

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3842979326004639x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

using namespace std;

int abssub(int a,int b){

	if(a>b)return a-b;

	return b-a;

}

int main(){

	int n,h[100001],bp[100001]={0};

	cin>>n; 

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

		cin>>h[i];

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

		if(i==2)bp[i]=bp[i-1]+abssub(h[i],h[i-1]);

		else{

			bp[i]=min(bp[i-1]+abssub(h[i],h[i-1]),bp[i-2]+abssub(h[i],h[i-2]));

		}

	}

	cout<<bp[n]<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3953719635154964x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);





}

int n;

vector<ll>v;

ll mem[(int)1e5 + 10];

ll solve(int i)

{

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

	ll& ret = mem[i];

	if (~ret)return ret;

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

	if (i + 2 < n)

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

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		clr(mem, -1);

		cin >> n;

	v = vector<ll>(n);

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

	ex(solve(0));

}

// Faster C++ solution for the same problem (measured speedup: 1.80038403437845x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

#include <sstream>

using namespace std;

#include <bits/stdc++.h>

#include<utility>

#include <algorithm>

#include<iterator>

#define  go return 0

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

#define rep(j,m) for(int j=0;j<m;++j)

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

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

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

#define F first

#define S second

#define pb push_back

#define pf push_front

#define mp make_pair

typedef long long ll;

typedef double db;

typedef vector<ll> ve;

typedef vector< vector <int> >vee;

typedef vector<pair<int,int> >vepa;

typedef set<char>se;

typedef map<ll,ll> mpp;

const ll N=1e5+5;

bool f;

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

ll solve(ll i=0)

{

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

  ll &ret=dp[i];

  if(~ret)return ret;

   return ret=min(solve(i+1)+abs(arr[i]-arr[i+1]) ,i+2<n? solve(i+2)+abs(arr[i]-arr[i+2]):12345 );

}

int main()

{

ios::sync_with_stdio(0);

cin.tie(0);

   cin>>n;

   memset(dp,-1,sizeof dp);

   rep(i,n)

    cin>>arr[i];

   cout<<solve();



go;

}


// Faster C++ solution for the same problem (measured speedup: 1.7940441197232946x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

using namespace std;



vector<long long int> dp(100005,-1);

long long int n,ara[100005];

int f(long long int i){



    if(i==n) return 0;



    if(i==n-1) return abs(ara[i+1]-ara[i]);



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



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

    return dp[i];





}









int main()

{

    long long int i;

    cin>>n;

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

        scanf("%lld",&ara[i]);

    }



    cout<<f(1)<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7936073097124479x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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



#define clr(x) memset((x), 0, sizeof(x))

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

#define pb push_back

#define mp make_pair

#define For(i, st, en) for(int i=(st); i<=(int)(en); i++)

#define Ford(i, st, en) for(int i=(st); i>=(int)(en); i--)

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

#define forn2(i, n) for(int i=0; i<=(int)(n); i++)

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

#define fori(it, x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)

#define in(x) int (x); input((x));

#define x first

#define y second

#define less(a,b) ((a) < (b) - EPS)

#define more(a,b) ((a) > (b) + EPS)

#define eq(a,b) (fabs((a) - (b)) < EPS)

#define remax(a, b) ((a) = (b) > (a) ? (b) : (a))

#define remin(a, b) ((a) = (b) < (a) ? (b) : (a))

#define salfinal(x) cout<<x<<endl;

using namespace std;

const int MAX =1e5+1;

int v[MAX];

int memo[MAX];

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()

{

    ios_base::sync_with_stdio(false);

    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: 1.4002842409575664x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

            *p = *p * 10 + ch - '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: 1.3871598412951247x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

#define ll long long

using namespace std;

#define inf 0x3f3f3f3f 

int main(){

	ll t , h[100001] , dp[100001] = {0};

    cin >> t;

	for(int i = 0 ; i < t ; ++i) scanf("%lld",h+i);

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

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

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

        cout<<dp[t-1];

}

// Faster C++ solution for the same problem (measured speedup: 1.8047185541463286x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

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

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define gg(x) getInt(&x)

using namespace std;

typedef long long ll;

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

ll n;

ll dp[maxn];

ll a[maxn];

int main()

{

    gbtb;

    cin>>n;

    repd(i,1,n)

    {

        cin>>a[i];

    }

    dp[1]=0;

    dp[0]=0;

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

    repd(i,3,n)

    {

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



    }

    cout<<dp[n];

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

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

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

        }

    }

    else {

        *p = ch - '0';

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

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

        }

    }

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF+5];

long long mem[INF];



long long rec(int pos)

{

    if(pos==n-1)

    {

        return 0;

    }

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

    long long tot=maxx;

    long long left=abs(a[pos]-a[pos+1]);

    long long right=abs(a[pos]-a[pos+2]);

    if(pos+1<n)

    {

        tot=min(tot,left+rec(pos+1));

    }

    if(pos+2<n)

    {

        tot=min(tot,right+rec(pos+2));

    }

    return mem[pos]=tot;

}



int main()

{

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

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=rec(0);

    printf("%lld\n",ans);

}


// Faster C++ solution for the same problem (measured speedup: 2.5938557845846235x):
//A 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e5+10;

const int INF = 1e9+10;

int raw[maxn],f[maxn];

int main(){

	int n;

	cin >> n;

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

		cin >> raw[i];

		f[i] = INF;

	}

	f[1] = 0;

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

		if(i-1 > 0)f[i] = min(f[i],f[i-1] + abs(raw[i] - raw[i-1]));

		if(i-2 > 0)f[i] = min(f[i],f[i-2] + abs(raw[i] - raw[i-2]));

	}

	cout << f[n];

	return 0;

}