// Slower C++ solution:
// Created by ahmed_drawy on 10/09/19.

///dp educational contest atcoder - vjudge

#include <bits/stdc++.h>

#include <string.h>

#pragma clang diagnostic push

#pragma ide diagnostic ignored "OCUnusedMacroInspection"

using namespace std;

#define lp(i,start , end)       for(int i = start ; i<end ; ++i)

#define lllp(i,start , end)       for(ll i = start ; i<end ; ++i)

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

#define Rlp(i,start , end)      for(int i = start ; i>end ; --i)

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

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

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

#define inf 0x3f3f3f3f

//#define dbg (v)    for(auto it : v)cout<<it<<' ';

typedef long long               ll;

typedef vector<int>             vi;

typedef vector  <ll>            vll;

typedef vector<vector<int> >    adj;

typedef pair<int ,int>          pii;

const double EPS =1e-7;

const int OO = 1e6;

#define endl '\n'

#define cint(c) (int)c -'0';

#define itc(c) (char)c +'A';

bool sortpair( const pair<int , int> & x, const pair<int , int> & y){

    return  x.first != y.first ? x.first < y.first : x.second > y.second ;

}

int dcomp(double x , double y){

    return fabs(x-y) <= EPS? 0: x>y?1:-1;

}



void smile() {

    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);

#ifndef ONLINE_JUDGE

//    freopen("/home/tw3/Desktop/Desktop/training/in.txt", "r", stdin);

//    freopen("/home/tw3/Desktop/Desktop/training/out.txt" , "w" , stdout);

#endif

}

#define fastin smile()



int n ;

const int N = 1e5 +3;

ll dp[N];

int h[N];

//ll solve(int indx , int state){

//    if(indx == n-1 )return 0 ;

//    ll &ret = dp[indx  ] [ state];

//    if(~ret )

//        return ret;

//    ret=  INT_MAX;

//    for (int i = 1; i <=2 && i+indx < n ; ++i) {

//        ret = min(ret , abs(h[indx]  - h[indx+ i])+solve(indx + i , i ));

//    }

//    return ret ;

//}

int main() {

    fastin;

    cin >> n;

    inN(h,  n );

    clr(dp , 0);

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

        if(j>=2){

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

        }

        else if(j ==1){

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

        }

    }

    cout << dp[n-1];

//

//    clr(dp , -1);

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

}

// Faster C++ solution for the same problem (measured speedup: 1.287705810170636x):
/**       In the name of Allah, the Gracious, the Merciful.

*               Praise be to Allah, Lord of the Worlds.

*               **************************************

*

*    I have not failed, I have just found 1000 WAYS that won't work.

*                 __________________________________

*                  Author  : Abd El-Twab  M. Fakhry

*                 __________________________________

*/



#include <ext/rope>

#include <bits/stdc++.h>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/assoc_container.hpp>



#define  print(arr, l, r)    for(int i = (l); i < (r); cout << arr[i++] << " \n"[i == r])

#define  read(arr, l, r)     for(int i = (l); i < (r); cin >> arr[i++])

#define  repr(i, r, l)       for (int i = r; i > l; i--)

#define  rep(i, l, r)        for (int i = l; i < r; i++)

#define  Unique(n)           (n).erase(unique(all(n)), (n).end())

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

#define  RT(n)               return cout << (n) << endl, 0;

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

#define  endl                '\n'

#define  D(x)                cout << #x << " = " << (x) << '\n'

#define  F                   first

#define  S                   second

#define  PB                  push_back

#define  EB                  emplace_back

#define  lsb(x)              (x & -x)

#define  get(n)              find_by_order(n)

#define  sqrt(n)             sqrtl(n)

#define  pow(a, b)           powl(a, b)

#define  indexOf(n)          order_of_key(n)

#define  isPowerOfTwo(x)     (x && !(x & (x - 1)))



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef long long            ll;

typedef unsigned long long   ull;



/**---------------------->>  Constants  <<-----------------------**/



const double    Pi      = 2 * acos(0.0),    Euler = 2.71828182845904523536,

                Epsln   = 1e-9,           GoldenR = (1 + sqrtl(5.0)) / 2;



const ll        MgcPrim = 2305843009213693951;

const int       MxsZ    = (int)1e5 + 9,        oo = 0x3f3f3f3f,

 		        ModB2   = (1 << 30) + 7,   ModBT7 = (int)1e9 + 7, ModBT = (int)1e9;



/**---------------->>  Functions & Shortcuts  <<-----------------**/



template <class T>

T Gcd(T a, T b) { return b ? Gcd(b, a % b) : a; }



template <class T>

T Lcm(T a, T b) { return a / Gcd(a, b) * 1ll * b; }



template <class T> /** Phi(a * b) = Phi(a) * Phi(b) where a and b Prime Factors and k their power **/

T Phi(T a, T k) { return pow(a, k - 1) * (a - 1); }



template <class T>

using indexed_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update >;



template <class T>

using matrix = vector < vector <T> >;



template <class T>

using min_heap = priority_queue < T, vector <T>, greater <T> >;



/** Watch **/



template <class T>

string ToString(T x) { stringstream sout; sout << x; return sout.str(); }



#ifdef BFS

int dx8[8] = { 1, -1, 0, 0, 1, 1, -1, -1 }, dx4[4] = { 0, 0, 1,-1};

int dy8[8] = { 0, 0, 1, -1, 1, -1, 1, -1 }, dy4[4] = { 1,-1, 0, 0};

#endif // BFS



/**------------------>>  Fast input / output  <<-----------------**/



void Fast()

{

    cin.sync_with_stdio(0);

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

    cin.exceptions(cin.failbit);

}



/**----------------->>  input / output Files  <<-----------------**/



void File()

{

    #ifdef FIO

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

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

   	#endif // FIO

}



/**------------------>>  Initialization gap  <<------------------**/



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



int Frog(int nx)

{

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

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

}



inline void Solve()

{

    memset(dp, -1, sizeof dp);



    cin >> n; read(a, 0, n);

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



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

}



/**------------------------->>  Main  <<-------------------------**/



int main()

{

    Fast(); File(); Solve();

}