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

#include <algorithm>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <vector>

#include <string>

#include <climits>

#include <map>

#include <set>

#include <stack>

#include <queue>

#include <deque>

#include <stdio.h>

//#define int long long

using namespace std;

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

#define endl '\n'

#define OnlineJudgec

#define F first

#define S second

#define  precision(n)  cout << fixed << setprecision(n)

#define clr(v,d ) memset(v, d, sizeof(v))

typedef long long ll;

typedef vector<int> vi;

void init()

{

    cin.tie(0);

    cin.sync_with_stdio(0);

    cout.tie(0);

    #ifdef OnlineJudge

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    #endif

}

ll S(ll x){ll s = 0;while(x){s += x % 10;x /= 10;}return s;} /// sum of digits.

ll powe(ll a , ll n){if(n == 1)return a;return a * powe(a ,n-1);}

ll com(int n,int m){ll z = 1;for(int i = 0; i < m;i++){z = z * ( n - i) / (i + 1) ;}return z;}

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

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

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

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

/// <<------------------------------------------------------------------------------>>

const int mx = 100009;

int n;

int arr[mx];

int dp[mx];

int solve(int idx)

{

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

    int &rst = dp[idx];

    if(~rst) return rst;

    int choice = 1e9 , choice2 = 1e9;

    if(idx + 1 < n)

        choice = solve(idx+1) + abs(arr[idx] - arr[idx + 1]);

    if(idx + 2 < n)

       choice2 = solve(idx+2) + abs(arr[idx] - arr[idx + 2]);

    return rst = min(choice , choice2);

}

/// <<------------------------------------------------------------------------------>>

int main()

{

    init();

    cin >> n;

    loop(n) cin >> arr[i];

    clr(dp,-1);

    cout<<solve(0);

}


// Faster C++ solution for the same problem (measured speedup: 1.4192526316682945x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[0]);

	return 0;

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



using namespace std;



#define F first

#define S second



int N;

int H[100005];

int dp[100005][5];



int solve(int cost, int ind, int j) {

	if (ind > N) return INT32_MAX;

	if (ind == N) return cost;



	if (~dp[ind][j]) return dp[ind][j];

	

	dp[ind][j] = cost;

	dp[ind][j] += min(solve(abs(H[ind] - H[ind + 1]), ind + 1, 0), 

					solve(abs(H[ind] - H[ind + 2]), ind + 2, 1));

	return dp[ind][j];

}



int main() {

	cin >> N; 

	H[0] = INT32_MAX; // to avoid RTE

	for (int i = 1; i <= N; i++) cin >> H[i];

	int ans = 0;

	

	memset(dp, -1, sizeof dp);

	cout << solve(0, 1, 0) << endl;



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.0201617900616666x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[0]);

	return 0;

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

using namespace std;

#define ll long long

#define ull unsigned long long

#define PII pair<int,int>

#define PSI pair<string,int>

#define PIII pair< int, pair<int,int> >

const int MXN = 1e5;

int a[MXN+5], dp[MXN+5];

int N,K;



int rec(int pos){



    if(pos==N) return 0;

    if(pos>=N) return 1<<30;

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

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

}



void solve(){



    cin>>N;

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

    for(int i=1; i<=N; i++) cin>>a[i];

    cout<< rec(1) << "\n";

}



int main(){



    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);

    solve();

}


// Faster C++ solution for the same problem (measured speedup: 1.4201867682010716x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[0]);

	return 0;

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

using namespace std;



int n;

int h[100005];

int dp[100005];



int solve(int x){

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

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

    else if (x == n-2){

        return dp[x] = abs(h[x+1]-h[x]);

    }

    else {

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

    }

}



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

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin >> n;

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

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

        cin >> h[i];

    }

    cout << solve(0) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4204723070634062x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[0]);

	return 0;

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

#define inf INFINITY

#define file freopen( "input.txt" , "r" , stdin) ; freopen( "output.txt" , "w" , stdout) ;

#define Gamed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define _3102_ main()

#define endl '\n'

typedef long long ll;

typedef unsigned long long ull;

typedef long double ld;

using namespace std;

const int N = 1e5 + 100;

const ll MX = 1e9;

const ll mod = 1e9 + 7;

const double eps = 1e-9 ;



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

ll solve( int i ){

	if ( i < 0 ) return 0 ;

	ll &ret = dp[i] ;

	if ( ~ret ) return ret ;

	if ( i == 0 ) return ret = 0 ;

	if ( i == 1 ) return ret = abs( h[1] - h[0]) ; 

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

}



int _3102_{

	Gamed

	cin >> n ;

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

		cin >> h[i] ;

	}

	memset ( dp , -1 , sizeof dp ) ;

	cout << solve ( n - 1 ) ;



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.8237558326013408x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[0]);

	return 0;

}
// 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: 1.8349065922167875x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[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: 1.841834445836632x):
#include <bits/stdc++.h>

using namespace std;

#define MAXN 100000



int h[MAXN];

int dp[MAXN];

int n;



int main() {

	int i;

	scanf("%d",&n);

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

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

	}

	dp[n-1]=0;

	dp[n-2]=abs(h[n-1]-h[n-2]);

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

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

	}

	

	printf("%d\n",dp[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: 1.3702968097341708x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// 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: 1.7614793182920747x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// 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: 4.963521896285241x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// 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: 1.3613089559988611x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// 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: 1.3702533549006901x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

const int mod = 1e9+9;



const int ppt = 1e5+3;

int n;

int dp[ppt];

int a[ppt];

int f(int i){

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

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

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

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

	return dp[i]=jpr;

}

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: 1.3702533549006901x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

const int mod = 1e9+9;



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: 1.3702533549006901x):
#include <bits/stdc++.h>

#define f first

#define s second

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

#define pb push_back

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

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

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

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

#define deb(x) cout << #x": " << (x) << endl;



int main(){ _

  lli n;

  cin >> n;

  vector <lli> a(n);

  lli x;

  fore (i,0,n){

    cin >> x;

    a[i] = x;

  }

  vector <lli> b(n);

  b[0] = 0;

  b[1] = abs(a[0] - a[1]) + b[0];

  fore (i,2,n){

    if ((abs(a[i - 1] - a[i]) + b[i - 1]) < (abs(a[i - 2] - a[i]) + b[i - 2])){

      b[i] = abs(a[i - 1] - a[i]) + b[i - 1];

    }

    else{

      b[i] = abs(a[i - 2] - a[i]) + b[i - 2];

    }

  }

  cout << b[n - 1] << ENDL;

  return 0;

}

// Hola, si estás viendo esto te notifico que te detuviste a leer mi código.