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

 

using namespace std;

 

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

#define pd(n) printf("%.12lf", n)

#define pf(n) printf("%d", n)

#define p() printf("\n")

#define ps() printf(" ")

#define ll long long

#define ull unsigned long long

#define MAX 1e9

const int mod = 1000 * 1000 * 1000 + 7;

#define ex 1e18

#define dbl 2e09

#define qu(q) queue<int> q

#define pqu(q) priority_queue<int> q

#define len(s) s.length()

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

#define pb(x) push_back(x)

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

#define rall(x) x.rbegin(), x.rend()

#define F first

#define S second

#define clr(x) x.clear()

#define LOG(x) std::cout << x << std::endl;

 

const int INF = 0x3f3f3f3f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const double PI = 3.14159265359;

 

int dx[6] = {-1,-1,0,0,1,1};

int dy[6] = {-1,0,-1,1,0,1};



int a[100005];

int dp[100005];

int n;

int solve(int idx){

    if(idx==(n-1))

        return 0;

    if(idx>=n)

        return INF;

    int &ret = dp[idx];

    if(~ret)

        return ret;

    ret = INF;

    ret=min(ret,solve(idx+1)+abs(a[idx+1]-a[idx]));

    ret=min(ret,solve(idx+2)+abs(a[idx+2]-a[idx]));

    return ret;

}







int main(){



    cin>>n;

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

        cin>>a[i];

    }

    memset(dp,-1,sizeof dp);

    cout<<solve(0)<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.357360286831826x):
#include <bits/stdc++.h>

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

#define ii          pair<int,int>

#define F           first

#define S           second

#define pb          push_back

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

#define rall(x)     (x).rbegin(),(x).rend()

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

#define forr(i,a)   for(int i = a; i >= 0; i--)

#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 

#define w(t)        while(t--)

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

#define cls(a,car)  memset(a,car,sizeof (a))

#define db(x)       cerr << #x << " is " << x << '\n'

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

#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"

#define angle(x)    double(x * acos(-1) / 180.0)

using namespace std;

typedef long long   ll;

typedef vector<int> vi;

typedef vector<ii>  vii;

const int N = 1 * 1e5 + 5;

const ll mod = 1e9 + 7;

const double E = 1e-9;

//int dx[] = {1, 0}, dy[] = {0, 1};

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

//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

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

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int v[N];

int dp[N];

int n, k = 2;

int f(int i){

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

    int &ans = dp[i];

    if(ans == -1){

        ans = 1e9;

        for(int j = 1; j <= k; j++)

            if(i + j < n)

                ans = min(ans, f(i + j) + abs(v[i] - v[i + j]));

            else

                break;

    }

    return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n;

    fore(i, 0, n)cin >> v[i];

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

        dp[i] = 1e9;

    for(int i = n - 1; i >= 0; i--){

        int &ans = dp[i];

        if(i == n - 1){

            ans = 0;

            continue;

        }

        ans = 1e9;

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

            if(i + j < n)

                ans = min(ans, dp[i + j] + abs(v[i] - v[i + j]));

            else

                break;

        }

    }

    

    cout << dp[0] << '\n';

    EjecuteTime;

    return 0;

}