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;
int vis[1000];
int Pow[5];
int change(string p) {
  int sum = 0;
  for (int i = 0; i < p.size(); i++) {
    sum += (p[i] - 48);
    sum *= 10;
  }
  return sum / 10;
}
string minstr(string s) {
  int i = 0, j = 1, k = 0;
  int len = s.size();
  while (j < len && i < len && k < len) {
    char tmp = s[(i + k) % len] - s[(j + k) % len];
    if (tmp == 0) {
      ++k;
    } else {
      if (tmp < 0) j += k + 1;
      k = 0;
      if (tmp > 0) i = i + k + 1;
      if (i == j) j++;
    }
  }
  int t = max(i, j);
  string p = s.substr(t) + s.substr(0, t);
  return p;
}
void init() {
  vis[0] = ' ';
  for (int i = 0; i < 26; i++) {
    vis[i + 1] = i + 'A';
  }
  vis[27] = '\'';
  vis[29] = '-';
  vis[28] = ',';
  vis[30] = '.';
  vis[31] = '?';
  Pow[0] = 1;
  for (int i = 1; i < 5; i++) {
    Pow[i] = Pow[i - 1] * 2;
  }
}
int a[100000];
int main() {
  int n;
  while (cin >> n) {
    int dex = 0;
    int minn = 9999999;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
      int sum = 0;
      for (int j = 1; j <= n; j++) {
        sum += ((abs(j - i) + (j - 1)) + (i - 1)) * a[j] * 2;
      }
      if (sum < minn) {
        minn = sum;
        dex = i;
      }
    }
    cout << minn << endl;
  }
  return 0;
}
