stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:


/**

    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;

}


// Faster C++ solution for the same problem (measured speedup: 1.8735043006726075x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 6.716504896079487x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 3.6292503276840082x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

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

using namespace std;

typedef long long ll;

ll dp[100009];

int arr[100009];

ll sum = 0;

long long Frog1(int n){

 

    if(n==0) return 0;

    else if(n==1) return abs(arr[n] - arr[n-1]);

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

    else {

            ll x = Frog1(n-1);

            ll y = Frog1(n-2);

        dp[n] = min(abs (arr[n-1] - arr[n]) + x,abs(arr[n-2] - arr[n]) + y);

        //cout << n << " = " << dp[n] << endl;

         return dp[n];

    }

}

 

 

int main()

{

    int n;

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

    cin >> n;

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

    cout << Frog1(n-1) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.2869026105846721x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

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

using namespace std;

#define ll long long

#define ld long double

#define sf scanf

#define pf printf

#define pb push_back

#define llu unsigned long long

#define U unsigned int

#define SIZE 1000002

#define pie 3.14159265358979323

#define minuss 1e-6

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

 

#define BIG 100000000

 

ll  lcm(ll a,ll b)

{

    ll GCD=__gcd(a,b);

    ll lcm=(a*b)/GCD;

    return lcm;

}

ll dp[100005];

ll weight[1000005],value[1000005];

ll total,val;

ll call(ll n)

{

    if(n>=total)

        return 0;

 

    if(dp[n]!=-1)

        return dp[n];

 

    ll res1=BIG,res2=BIG;

 

   // if(weight[n]+c<=val)

    res1=abs(weight[n]-weight[n+1]) + call(n+1);

    if(n+2<=total) res2=abs(weight[n]-weight[n+2]) + call(n+2);

    //cout<<n<<" "<<res1<<" "<<res2<<endl;

    dp[n]=min(res1,res2);

    return dp[n];

 

}

int main()

{

    cin>>total;

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

        cin>>weight[i];

 

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

    ll ret=call(1);

    cout<<ret<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2864201114465614x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

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

using namespace std;

int n;

long long h[100005];

long long dp[100005];



long long solve(int i){

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

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



	dp[i] = 1e17;

	if(i+1 <= n-1)

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

	if(i+2 <= n-1)

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



	return dp[i];

}



int main(){

	scanf("%d", &n);

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

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

	memset(dp, -1, sizeof dp);

	cout<<solve(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.286843168295933x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

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

using namespace std;

#define ll long long

#define pb push_back

#define mp make_pair

#define ull  unsigned long long

#define vll  vector <long long>

#define pll pair <long long,long long>

#define f first

#define s second

#define up upper_bound

#define lp lower_bound

#define pq priority_queue

#define inf 1e10

#define minf -1e10

#define pi 3.14159265

#define mod 300000000000000000

#define fastio ios_base::sync_with_stdio(0);cin.tie(NULL);

int main()

{

    ll n;

    cin>>n;

    vll vec;

    ll ara[100001];

    memset(ara,0,sizeof(ara));

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

    {

        ll num;

        cin>>num;

        vec.pb(num);

    }

    ll ans=0;

    ara[n-2]=abs(vec[n-2]-vec[n-1]);

    for(ll i=n-3;i>=0;i--)

    {

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

    }

    cout<<ara[0]<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.295610844505381x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long



int ara[100000];

int dp[100000];



int f(int i, int n)

{

    if(i == n) return 0;

    if(dp[i] != -1)

        return dp[i];

    int cost = f(i+1,n) + abs(ara[i] - ara[i+1]);

    if(i+2<=n)

        cost = min(cost, f(i+2,n)+abs(ara[i] - ara[i+2]));

    dp[i] = cost;



    return cost;

}



int main()

{

   int n;

   cin>>n;



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

   {

       cin>>ara[i];

   }



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

   cout<<f(1,n)<<endl;



   return 0;

}
// Slower C++ solution:
#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));

}


// Faster C++ solution for the same problem (measured speedup: 1.2750994964940932x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

}
// Slower C++ solution:


/**

    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;

}


// Faster C++ solution for the same problem (measured speedup: 1.8699282610124166x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 6.703684809209323x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

}
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 3.622323018734731x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

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

using namespace std;

typedef long long ll;

ll dp[100009];

int arr[100009];

ll sum = 0;

long long Frog1(int n){

 

    if(n==0) return 0;

    else if(n==1) return abs(arr[n] - arr[n-1]);

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

    else {

            ll x = Frog1(n-1);

            ll y = Frog1(n-2);

        dp[n] = min(abs (arr[n-1] - arr[n]) + x,abs(arr[n-2] - arr[n]) + y);

        //cout << n << " = " << dp[n] << endl;

         return dp[n];

    }

}

 

 

int main()

{

    int n;

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

    cin >> n;

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

    cout << Frog1(n-1) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.2844462432453487x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

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

using namespace std;

#define ll long long

#define ld long double

#define sf scanf

#define pf printf

#define pb push_back

#define llu unsigned long long

#define U unsigned int

#define SIZE 1000002

#define pie 3.14159265358979323

#define minuss 1e-6

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

 

#define BIG 100000000

 

ll  lcm(ll a,ll b)

{

    ll GCD=__gcd(a,b);

    ll lcm=(a*b)/GCD;

    return lcm;

}

ll dp[100005];

ll weight[1000005],value[1000005];

ll total,val;

ll call(ll n)

{

    if(n>=total)

        return 0;

 

    if(dp[n]!=-1)

        return dp[n];

 

    ll res1=BIG,res2=BIG;

 

   // if(weight[n]+c<=val)

    res1=abs(weight[n]-weight[n+1]) + call(n+1);

    if(n+2<=total) res2=abs(weight[n]-weight[n+2]) + call(n+2);

    //cout<<n<<" "<<res1<<" "<<res2<<endl;

    dp[n]=min(res1,res2);

    return dp[n];

 

}

int main()

{

    cin>>total;

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

        cin>>weight[i];

 

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

    ll ret=call(1);

    cout<<ret<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2839646650744614x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

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

using namespace std;

int n;

long long h[100005];

long long dp[100005];



long long solve(int i){

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

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



	dp[i] = 1e17;

	if(i+1 <= n-1)

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

	if(i+2 <= n-1)

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



	return dp[i];

}



int main(){

	scanf("%d", &n);

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

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

	memset(dp, -1, sizeof dp);

	cout<<solve(0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2843869144167077x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

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

using namespace std;

#define ll long long

#define pb push_back

#define mp make_pair

#define ull  unsigned long long

#define vll  vector <long long>

#define pll pair <long long,long long>

#define f first

#define s second

#define up upper_bound

#define lp lower_bound

#define pq priority_queue

#define inf 1e10

#define minf -1e10

#define pi 3.14159265

#define mod 300000000000000000

#define fastio ios_base::sync_with_stdio(0);cin.tie(NULL);

int main()

{

    ll n;

    cin>>n;

    vll vec;

    ll ara[100001];

    memset(ara,0,sizeof(ara));

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

    {

        ll num;

        cin>>num;

        vec.pb(num);

    }

    ll ans=0;

    ara[n-2]=abs(vec[n-2]-vec[n-1]);

    for(ll i=n-3;i>=0;i--)

    {

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

    }

    cout<<ara[0]<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 1.293137855378822x):
#include <bits/stdc++.h>



#define TR 100000



int l[TR];

using namespace std;

vector<int> h;





int solve(int p) {

	if(p == 0)

		return 0;

	if(p < 0)

		return 10000000;

	if(l[p] != -1)

		return l[p];

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

	return l[p];



}



int main() {

	int n;

	cin >> n;

	memset(l, -1, TR * sizeof(int));

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

		cin >> tmp;

		h.push_back(tmp);

	}

	solve(n - 1);

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

}