stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int h[N];

int mem[N];



int go(int idx){

    if(idx >= n)

        return 0;

    if(mem[idx] != -1)

        return mem[idx];

    int best = 1e9;

    if(idx + 1 <= n)

        best = go(idx + 1) + abs(h[idx] - h[idx + 1]);

    if(idx + 2 <= n)

        best = min(best , go(idx + 2)  + abs(h[idx] - h[idx + 2]));

    return mem[idx] = best;

}



int main(){

    scanf("%d" , &n);

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

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

    }

    memset(mem , -1 , sizeof mem);

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

}

//1 2 3


// Faster C++ solution for the same problem (measured speedup: 1.3884315614205816x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

#define pii pair<int,int>

#define vi vector<int>

#define ppii pair<pii>



using namespace std;

const int maxn=1e5+5;



int n;

vi arr;

int dp[maxn];



int dfs(int pos)

{

    if(pos==n){return 0;}



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



    int ans=(1<<29);



    if(pos<n)

        ans=min(ans,dfs(pos+1) + abs(arr[pos]-arr[pos+1]));

    if(pos<n-1)

        ans=min(ans,dfs(pos+2) + abs(arr[pos]-arr[pos+2]));

    dp[pos]=ans;



    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

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



    cin>>n;



    arr=vi(n+1);



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

        cin>>arr[i];



    memset(dp,-1,sizeof dp);



    cout<<dfs(1);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4150331631742163x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

#include <queue>

typedef long long ll;



using namespace std;



const int N=1e5+5;





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

   int n,k=2;

   cin>>n;

   int arr[N]={},dp[N]={};

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

    cin>>arr[i];

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

   dp[0]=0;

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

   {



     for(int j=1;j<=k;j++)

     {

         if(i-j>=0)

         {

             if(dp[i]==-1)

             {

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

             }

             else

             {

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





             }





         }



     }









   }

   cout<<dp[n-1];



   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8627831620705422x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;



typedef long long ll;

#define mp make_pair

#define f first

#define s second

#define pb push_back



const int inf = 1e7;

const int N  = 1e5 + 5;



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

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



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



int solve(int idx) {



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



  int &ret = dp[idx];



  if(ret!=-1) return ret;



  ret = solve(idx+1) + abs(h[idx]-h[idx+1]);



  if(idx!=n-2)

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



  return ret;



}



int main() {



    ios_base::sync_with_stdio(false);

    cin.tie(0);



    cin >> n;



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



      cin >> h[i];



    }



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



    cout << solve(0) << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4140366507928976x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    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: 1.4007206053735601x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;

int a,b,c,d,dp[100000],cost[100000],weight[100];



void func(int a)

{

    dp[1]=0;

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

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

    {

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

    }

}



int main()

{

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

    cin>>a;

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

    {

        cin>>cost[i];

    }

    func(a);

    cout<<dp[a]<<endl;

    //cout<<nCr(a,b)<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 1.3960370498460373x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {



    cin.tie(0);

    cin.sync_with_stdio(0);

}



const int N = 1e5 + 9;

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



int sol(int i){

    if(i<0) return 1e9;

    if(i==0) return 0;



    if(~dp[i]) return dp[i];



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

}





int main() {



    cin >> n;



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



    memset(dp, -1, sizeof dp);



    cout << sol(n - 1);

}

// Faster C++ solution for the same problem (measured speedup: 1.3989163001744922x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

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

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n,arr[N];

int dp[N];

ll solve(int i){

    if(i>=n)return 1e15;

    if(i==n-1){

        return 0;

    }



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



    ll path1=abs(arr[i]-arr[i+1])+solve(i+1);

    ll path2=abs(arr[i]-arr[i+2])+solve(i+2);

    return dp[i]= min(path1,path2);

}

int main()

{

    memset(dp,-1,sizeof dp);

    scanf("%d",&n);

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

    cout<<solve(0)<<endl;

}




// Faster C++ solution for the same problem (measured speedup: 1.395110613726097x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

#define	       MEM(a,b)         memset(a,(b),sizeof(a))

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

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

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

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

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

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

#define		gcd(a,b)        __gcd(a,b)

#define		lcm(a,b)        (a*b)/gcd(a,b)

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

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



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

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

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

}