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.
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) {
    in >> i;
  }
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &i : v) {
    out << i << ' ';
  }
  return out;
}
inline void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  cout << fixed << setprecision(15);
}
long long bin_pow(long long a, long long b) {
  return b == 0 ? 1 : (b % 2 ? bin_pow(a, b - 1) * a : bin_pow(a * a, b / 2));
}
bool prime(long long n) {
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) return false;
  return true;
}
const int INFI = 1e9;
int main() {
  io();
  long long n, mini = INFI;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  for (int i = 0; i < n; ++i) {
    long long sum = 0;
    for (int j = 0; j < n; ++j) {
      sum += a[j] * 2 * (i + j + (abs(i - j)));
    }
    mini = min(mini, sum);
  }
  cout << mini;
}
