// Slower C++ solution:
/**       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], k;



int Frog(int nx)

{

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



    int cur = oo;

    rep(i, 1, k + 1)

    {

        if(nx - i >= 0)

            cur = min(cur, Frog(nx - i) + abs(a[nx] - a[nx - i]));

    }



    return dp[nx] = cur;

}



inline void Solve()

{

    memset(dp, -1, sizeof dp);



    cin >> n; read(a, 0, n); k = 2;

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



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

}



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



int main()

{

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

}


// Faster C++ solution for the same problem (measured speedup: 1.417572108345327x):
#include<bits/stdc++.h>

using namespace std;

typedef		long long       ll;

typedef         long double      dl;

typedef     unsigned long long  ull;

#define		pb              push_back

#define		PB              pop_back

#define		nn              "\n"

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

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

#define		zz(v)           (ll)v.size()

#define		ss              ' '

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

#define	       CLR(p)           memset(p,0,sizeof(p))

#define f0(i,b) for(int i=0;i<(b);i++)

#define f1(i,b) for(int i=1;i<=(b);i++)

#define f2(i,a,b) for(int i=(a);i<=(b);i++)

#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define rep(i,a,b,c) for(int i=(a);i!=(b);i+=(c))

#define         arrsize(a)      (sizeof(a)/sizeof(a[0]))

//#define       arrsize(a)      (sizeof(a)/sizeof(*a))

#define 	S(a)            scanf("%lld",&a)

#define 	SS(a,b)         scanf("%lld %lld",&a,&b)

#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)

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

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

#define		pi              acos(-1.0)

#define	       ff               first

#define	       sc              second

typedef        pair <int, int>   pii;

typedef       pair <ll, ll>      pll;

typedef      vector< pair <ll, ll> >      vpll;

typedef		vector<ll>      vll;

typedef		map<string,ll>      msl;

typedef		map<ll,ll>      mll;

#define yes cout << "YES\n"

#define no cout<<"NO\n"

//memset(ar,-1,sizeof(ar));

//#define sort(x) sort(x.begin(), x.end())

//sort(a,a+n,greater<ll>())

//for (auto it = mp.begin(); it != mp.end(); ++it){}

// string x(w.size(),'1');

#define MAX 1000005

#define precision(a) fixed << setprecision(a)



#define mod 1000000007



ll ar[MAX],ans[MAX];



int main()

{

 //O_O ;

   ll n;

   cin>>n;

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

    cin>>ar[i];

   ans[n]=0,

   ans[n-1]=abs(ar[n-1]-ar[n]);

   for(ll i=n-2;i>=1;i--)

   {

       ans[i]=min((abs(ar[i]-ar[i+1])+ans[i+1]),(abs(ar[i]-ar[i+2])+ans[i+2]));

      // cout<<<<ss<<i<<nn;

       // cout<<(abs(ar[i]-ar[i+2])+ans[i+2])<<ss<<i<<nn;



   }

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

   // cout<<ans[i]<<ss;cout<<nn;

   cout<<ans[1]<<nn;



    return 0;

}