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: 1.455891709907078x):
#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;

}
// 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: 5.219365546287692x):
#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;

}
// 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: 2.820273998493495x):
#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;

}
// 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: 1.7689804134225366x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



}
// 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: 5.025347817052502x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



}
// Slower C++ solution:


#include <iostream>

#include <bits/stdc++.h>





using namespace std;





int a[100005], p[100005],n, i;

int respuesta(int x)

{

    if(x==n) return 0;

    if(x>n) return 1e9;

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

    return p[x]=min(abs(a[x+1]-a[x])+respuesta(x+1), abs(a[x+2]-a[x])+respuesta(x+2));

}





int main()

{

    memset(p,-1,sizeof p);

    cin>>n;

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

    {

        cin>>a[i];

    }

    cout<<respuesta(1)<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3876170588186736x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



}
// 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: 1.3845585580964677x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



}
// 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: 1.7819474108824627x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



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

using namespace std;



int num[100001], memo[100001];



int dp(int n){

    if(n==0) return 0;

    if(n==1) return abs(num[0]-num[1]);

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

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

}



int main(){

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

    ios_base::sync_with_stdio(0);

    cin.tie(0);

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

    int n;

    cin>>n;

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

    cout<<dp(n-1)<<"\n";

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3941518655750318x):
#include <bits/stdc++.h>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

using namespace std;

const ll mod = 1e9 + 7;







//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

//		fact[i] = (i * fact[i - 1]) % mod;

//		inv[i] = modPower(fact[i], mod - 2);

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

//	return ((fact[n] * inv[n - r]) % mod);

//}

//

//ll ncr(ll n, ll r) {

//	return ((((fact[n] * inv[n - r]) % mod) * inv[r]) % mod);

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}





const ll N = 1e5 + 7;



ll x;

ll a[N] = {};

ll dp[N] = {};



int main()

{

    MohamedMotaz;

    

    cin >> x;



    f(i, 0, x) cin >> a[i];

    dp[0] = 0;

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

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

    cout << dp[x - 1];



   



}