// Original C++ solution (slower):
#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <functional>

#include<bits/stdc++.h>

using namespace std;

using namespace __gnu_pbds;





typedef tree<int, null_type, less_equal<int>, rb_tree_tag,

             tree_order_statistics_node_update>

    ordered_set;





#define pf                  printf

#define sf(a)               scanf("%d",&a)

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

#define sff(a,b)            scanf("%d %d",&a,&b)

#define sffl(a,b)           scanf("%lld %lld",&a,&b)1073. Square Country

#define sfff(a,b,c)         scanf("%d %d %d",&a,&b,&c)

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

#define sffff(a,b,c,d)      scanf("%d %d %d %d",&a,&b,&c,&d)

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

#define sfffff(a,b,c,d,e)   scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)

#define sfffffl(a,b,c,d,e)  scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)

#define sfc(a)              scanf("%c",&a)

#define pii                 pair<int,int>

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

#define pb(a)               push_back(a)

#define pbp(a,b)            push_back({a,b})

#define db                  double

#define ft                  float

#define ll                  long long

#define ull                 unsigned long long

#define pii                 pair<int,int>

#define ff                  first

#define ss                  second

#define sz(x)               x.size()

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

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

#define max3(a, b, c)       max(a, b) > max(b, c) ? max(a, b) : max(b, c)

#define min3(a, b, c)       min(a, b) < min(b, c) ? min(a, b) : min(b, c)

#define for0(i,n)           for(int i=0;i<n;i++)

#define for1(i,n)           for(int i=1;i<=n;i++)

#define forrev(i,n)         for(int i=n-1; i>=0; i--)

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

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

#define stlloop(x)          for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)

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

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

#define case1(z)            cout<<"Case "<<z<<": "

#define case2(z)            printf("Case %d: ",z)

#define PI                  acos(-1) //3.14159265358979323846264338328

#define valid(tx,ty)        tx>=0 && tx<row && ty>=0 && ty<col

#define intlim              2147483648

#define N                   200005

#define inf                 100000008

#define mod                 1000000007



/*------------------------------Graph Moves----------------------------*/

//const int fx[]={+1,-1,+0,+0};

//const int fy[]={+0,+0,+1,-1};

//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move

//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move

//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move

//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move

/*---------------------------------------------------------------------*/









ll a[N];

ll dp[N];







int main()

{

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

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

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

    ll n,i,x,y;

    cin>>n;

    for1(i,n)

    cin>>a[i];



    dp[1]=0;

    dp[2]=abs(a[1]-a[2]);

    for(i=3;i<=n;i++)

    {

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

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

        dp[i]=min(x,y);

    }

    cout<<dp[n]<<endl;

	return 0;

}






















































// Optimized solution to the same problem (faster):
#include<cstdio>

#include<algorithm>

#include<cstdlib>

using namespace std;

const int maxn=100005;

int cost[maxn],dp[maxn];

int main(){

    int n;

    scanf("%d",&n);

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

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

    }

    dp[2]=abs(cost[2]-cost[1]);

    dp[3]=min(abs(cost[3]-cost[1]),dp[2]+abs(cost[3]-cost[2]));

    for(int i=4;i<=n;i++){

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

    }

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

    return 0;

}

// Fastest known solution for this problem: