The Fair Nut lives in n story house. a_i people live on the i-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening. 

It was decided that elevator, when it is not used, will stay on the x-th floor, but x hasn't been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm: 

  * Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger. 
  * Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor). 
  * Moves from the b-th floor back to the x-th. 

The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |a - b| units of electricity.

Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the x-th floor. Don't forget than elevator initially stays on the x-th floor. 

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of floors.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

Examples

Input

3
0 2 1


Output

16

Input

2
1 1


Output

4

Note

In the first example, the answer can be achieved by choosing the second floor as the x-th floor. Each person from the second floor (there are two of them) would spend 4 units of electricity per day (2 to get down and 2 to get up), and one person from the third would spend 8 units of electricity per day (4 to get down and 4 to get up). 4 ⋅ 2 + 8 ⋅ 1 = 16.

In the second example, the answer can be achieved by choosing the first floor as the x-th floor.
// header files
#include<bits/stdc++.h>
#include<iomanip>
#include<stdio.h>
#include<cmath>

using namespace std;

//typedef
typedef string STR;
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef priority_queue<int> PQI;
typedef vector<LL> VLL;
typedef vector<ULL> VULL;
typedef vector<STR> VS;
typedef vector<char> VC;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef pair<STR,int> PSI;
typedef pair<int,STR> PIS;
typedef vector<PII> VII;
typedef map<int,int> MII;
typedef map<LL,LL> MLL;
typedef map<STR,int> MSI;
typedef map<char,int> MCI;
typedef map<int,STR> MIS;
typedef set<STR> SS;
typedef set<int> SI;
typedef set<LL> SLL;

//symbolic const
#define MOD 1000000007
#define MAXN 1e9
#define DIGITS_IN(i) (LL)log10(i)+1 // gives no of digits in a number



//STL keywords
#define F first
#define S second
#define PB push_back
#define PF push_front
#define MP make_pair

//functions
#define dec(n) cout<<fixed<<setprecision(n);
#define f(i, n)   for(LL i=0; i<n; i++)
#define fa(i,a,n) for(LL i=a; i<=n; i++)
#define fs(i, s) for(LL i=0; s[i]!='\0'; i++)

//IO
#define file freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define sync ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
LL maxLL(LL a, LL b){return (a>b?a:b);}
LL minLL(LL a, LL b){return (a<b?a:b);}

void solve(){
  int n, m=0, pos=0;
  cin>>n;
  int a[n], de[n]={0};
  f(i, n){
    cin>>a[i];
    if(a[i]>m){
      m=a[i];
      pos=i;
    }
  }
  f(i, n){
    de[i]=abs(pos-i);
  }
  int sum=0;
  f(i, n){
    sum+=a[i]*(i+de[i]+pos);
  }
  cout<<2*sum<<"\n";
}

int main() {
  //file
  sync;
  solve();
  return 0;
}
