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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

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

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

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

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e5+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

//==================================================================================================================================================================//



int n, h[N], dp[N];



int main() {

   scanf("%d", &n);

   ms(dp, inf); dp[n] = 0;

   for1(i, n) scanf("%d", h+i);

   for(int i = n; i; --i) {

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

      else {

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

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

      }

   }

   printf("%d\n", dp[1]);

   return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3708866874429355x):
#include <bits/stdc++.h>



#define cn count

#define pb push_back

#define in insert

#define fs first

#define sc second

#define sz size

#define len length

#define forn(i, n) for (ll i = 0; i < (ll) (n); ++i)

#define forni(i, n) for (ll i = 1; i < (ll) (n); ++i)

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



using namespace std;

using ll = long long;

const ll inf = 1e9+7; 

const double PI =  3.141592653589793;



int main(){

	IO;

	ll n;

	cin>>n;

	ll dp[n],h[n];

	dp[0]=0;

	forn(i,n){

		cin>>h[i];

	}

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

	for(ll i= 2;i<n;i++){

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

	}

	cout<<dp[n-1]<<'\n';

	return 0;

}





// _   .-')                .-. .-')                 .-')                                               

//( '.( OO )_              \  ( OO )               ( OO ).                                             

// ,--.   ,--.).-'),-----. ,--. ,--.  .-'),-----. (_)---\_) ,--. ,--.     .-----.  ,-.-')  .-'),-----. 

// |   `.'   |( OO'  .-.  '|  .'   / ( OO'  .-.  '/    _ |  |  | |  |    '  .--./  |  |OO)( OO'  .-.  '

// |         |/   |  | |  ||      /, /   |  | |  |\  :` `.  |  | | .-')  |  |('-.  |  |  \/   |  | |  |

// |  |'.'|  |\_) |  |\|  ||     ' _)\_) |  |\|  | '..`''.) |  |_|( OO )/_) |OO  ) |  |(_/\_) |  |\|  |

// |  |   |  |  \ |  | |  ||  .   \    \ |  | |  |.-._)   \ |  | | `-' /||  |`-'| ,|  |_.'  \ |  | |  |

// |  |   |  |   `'  '-'  '|  |\   \    `'  '-'  '\       /('  '-'(_.-'(_'  '--'\(_|  |      `'  '-'  '

// """    """     """""""" """  """"      """""""  """""""  """"""""      """"""   """         """""""