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



using namespace std;



const int N=1e5;

int n;

int arr[N];

int mem[N];



int solve(int i){

    if(i==n-1)

        return 0;

    if(mem[i]!=-1) return mem[i];

    int op1=solve(i+1)+abs(arr[i]-arr[i+1]);

    int op2=i+2<=n-1?solve(i+2)+abs(arr[i]-arr[i+2]):1000000000;

    return mem[i]=min(op1,op2);

}



int main() {

    scanf("%d",&n);

    for(int i=0;i<n;++i)

        scanf("%d",arr+i);

    memset(mem,-1,sizeof mem);

    printf("%d",solve(0));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3899935420024663x):
/* In the name of ALLAH */

#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<string>



// MAXIMUM and MINIMUM elements

#define mx_ari(a)   *max_element(a , a+sizeof(a)/4)

#define mn_ari(a)   *min_element(a , a+sizeof(a)/4)

#define mx_arl(a)   *max_element(a , a+sizeof(a)/8)

#define mn_arl(a)   *min_element(a , a+sizeof(a)/8)

#define mx_vec(v)   *max_element(v.begin() , v.end())

#define mn_vec(v)   *min_element(v.begin() , v.end())

// COUNT the occurence of an element

#define cnt_ari(a,b)   count(a , a+sizeof(a)/4 , b)

#define cnt_arl(a,b)   count(a , a+sizeof(a)/8 , b)

#define cnt_vec(v,b)   count(v.begin() , v.end() , b)

// INPUTs

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

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

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

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

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

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

#define scf(a)        scanf("%f",&a)

#define scff(a,b)     scanf("%f %f",&a,&b)

#define scfff(a,b,c)  scanf("%f %f %f",&a,&b,&c)

#define scd(a)        scanf("%ld",&a)

#define scdd(a,b)     scanf("%lf %lf",&a,&b)

#define scddd(a,b,c)  scanf("%lf %lf %lf",&a,&b,&c)

// OUTPUTs

#define pri(a)        printf("%d ",a)

#define prii(a,b)     printf("%d %d ",a,b)

#define priii(a,b,c)  printf("%d %d %d ",a,b,c)

#define prl(a)        printf("%lld ",a)

#define prll(a,b)     printf("%lld %lld ",a,b)

#define prlll(a,b,c)  printf("%lld %lld %lld ",a,b,c)

#define prd(a)        printf("%.4lf ",a)

#define prdd(a,b)     printf("%.4lf %.4lf ",a,b)

#define prddd(a,b,c)  printf("%.4lf %.4lf %.4lf ",a,b,c)

#define prf(a)        printf("%.4f ",a)

#define prff(a,b)     printf("%.4f %.4f ",a,b)

#define prfff(a,b,c)  printf("%.4f %.4f %.4f ",a,b,c)

// LOOPs

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

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

#define fod(i,n)      for(double i=0;i<n;i++)

#define fof(i,n)      for(float i=0;i<n;i++)

// NEWLINEs

#define nwl           << "\n"

#define nl            printf("\n");

// PAIRs and PUSHbacks

#define mp            make_pair

#define pb            emplace_back

// IMPORTANTs

#define ll          long long

#define ull         unsigned long long int

#define mod         1000000007

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



using namespace std;





int cost[100010];

int chi(int i , int n , int a[]){

   if(i < 0){

      return cost[0];

   }



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

   return chi(i-1 , n , a);

}



int main()

{

   int n;

   cin >> n;

   int a[n];

   foi(i,n){

      cin >> a[i];

   }



   cost[n-1] = 0;

   cost[n-2] = abs(a[n-1] - a[n-2]);



   cout << chi(n-3 , n , a);



   return 0;

}