// Original C++ solution (slower):
#include<iostream>

#include<vector>

#include<algorithm>

#include<queue>

#include<map>

#include<stack>

#include<cmath>

#include<iomanip>

#include<set>

#include<numeric>

#include<sstream>

#include<random>

#include<cassert>

#include<climits>

#include<cstring>

using namespace std;

typedef long long ll;

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

#define rrep(i, st, n) for (int i = st; i < n; ++i)

using pii = pair<int, int>;

const int inf = 1e9 + 7;

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

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

#define ceil(a, b) a / b + !!(a % b)

#define chmax(a, b) a = max(a, b)

#define chmin(a, b) a = min(a, b)



int n;

int h[100001];

int dp[100001];



int main() {

    cin >> n;

    rep(i, n) cin >> h[i];

    rep(i, 100001) dp[i] = inf;

    dp[0] = 0;

    rep(i, n - 2) {

        chmin(dp[i + 1], dp[i] + abs(h[i] - h[i + 1]));

        chmin(dp[i + 2], dp[i] + abs(h[i] - h[i + 2]));

    }

    chmin(dp[n - 1], dp[n - 2] + abs(h[n - 2] - h[n - 1]));

    cout << dp[n - 1] << endl;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

//type

typedef long long ll;

#define pii pair<int, int>

#define vi vector<int>

//x * y * 1.0 can cause overflow

//constant

#define inf (int)(1e9+7)

#define mod (ll)(1e9+7)

#define eps 1e-10

//omission

#define eb emplace_back

#define F first

#define S second

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

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

#define ios cin.tie(0); ios::sync_with_stdio(false)

#define FIX(a) cout << fixed << setprecision(a)

//loop

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

#define rep3(i, st, n) for (int i = st; i < n; ++i)

//manip

#define chmax(a, b) a = max(a, b)

#define chmin(a, b) a = min(a, b)

#define UNIQUE(v) v.erase(unique(v.begin(), v.end(), v.end());

#define fill(x, y) memset(x, y, sizeof(x))

#define ceil(a, b) a / b + !!(a % b)

template<class T> T power_mod(T a, T b) {return b ? power_mod(a * a % inf, b / 2) * (b % 2 ? a : 1) % inf : 1;}

template<class T> T power(T a, T b) {return b ? power(a * a, b / 2) * (b % 2 ? a : 1) : 1;}

#define LB(v, x) (int)(lower_bound(v.begin(), v.end(), x) - v.begin())

#define UB(v, x) (int)(upper_bound(v.begin(), v.end(), x) - v.begin())

template<typename T> T gcd(T a, T b) {if (b == 0) return a; return gcd(b, a % b);}

template<typename T> T lcm(T a, T b) {return a / gcd(max(a, b), min(a, b)) * b;} //先に割る!!

#define outa(a, n) rep(i, n) cout << a[i] << " "; cout << endl

#define outaa(a, m, n) rep(i, m) {rep(j, n) cout << a[i][j] << " "; cout << endl;} cout << endl

#define out(a) cout << a << endl

#define outv(v) rep(i, SZ(v)) cout << v[i] << " "; cout << endl



//algorithm

//double pointer, l start, how many adds, can be 0 -> init r = l, sum = 0

//bfs, not x, y, i, j

//not !(i % 2), i % 2 == 0

//minimal required terms

//cast caution

//look constraints always

//product of minimum 9 prime numbers is above 1e9





int main() {

    int n; cin >> n;

    int h[n]; rep(i, n) cin >> h[i];

    

    //sumではなくmin

    

    int dp[n]; fill_n(dp, n, 0);

    dp[0] = 0;

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

    dp[2] = min(abs(h[2] - h[0]), dp[1] + abs(h[2] - h[1]));

    rep3(i, 3, n) {

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

    }

    out(dp[n - 1]);

}


// Fastest known solution for this problem: