// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

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

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

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

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

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

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

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

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





}

int n;

vector<ll>v;

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

ll solve(int i)

{

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

	ll& ret = mem[i];

	if (~ret)return ret;

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

	if (i + 2 < n)

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

	return ret;

}

int main()

{

	Mohammed_Atef_Hassan

		clr(mem, -1);

		cin >> n;

	v = vector<ll>(n);

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

	ex(solve(0));

}

// Faster C++ solution for the same problem (measured speedup: 1.2850541930456818x):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef double dd;

typedef vector<int> vi;

typedef vector<ll> vl;

typedef pair<int,int> pii;



const int mx=1e5+11;

const double PI = acos(-1);

#define MOD 1000000007





#define nl '\n'

#define pb push_back

#define F first

#define S second

#define I insert

#define mp make_pair



#define sz(x) x.size()

#define sa(v) sort(v.begin(),v.end())

#define sd(v) sort(v.begin(),v.end(),greater<>())

#define rev(s) reverse(s.begin(),s.end())

#define tt int t; cin>>t; while(t--)

#define lo(i,a,b) for(ll i=a;i<b;i++)

#define be(a) (a).begin(),(a).end()

#define rbe(a) (a).rbegin(),(a).rend()



#define sqr(a) ((a) * (a))

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

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

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



#define yes cout<<"YES"<<nl;

#define no cout<<"NO"<<nl;



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

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

#define frac cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);



#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

void faltu ()

{

    cerr << endl;

}

template < typename T, typename ... hello>void faltu( T arg, const hello &... rest)

{

    cerr << arg << ' ';

    faltu(rest...);

}



int dp[mx],n,a[mx];



int solve (int i)

{

    if ( i >= n )

        return 0;

    if ( dp[i]!= -1 )

        return dp[i];



    int ret1 = INT_MAX, ret2 = INT_MAX;



    if (i+1<=n )

        ret1 = abs(a[i]-a[i+1])+ solve (i+1);



    if(i+2<=n)

        ret2 = abs(a[i]-a[i+2])+solve (i+2);

    ///dbg(i);

    dp[i]= min ( ret1, ret2 );



    return dp[i];

}



int main()

{

    su;



    mem ( dp, -1 );



    cin>>n;



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

    {

        cin >>a[i];

    }



    cout << solve( 1 )<< nl;





    return 0;

}