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;
const int MOD = (int)1e9 + 7;
const int oo = (int)1e9;
const long long OO = (long long)1e18;
const double PI = acos(-1.0);
const int N = (int)1e5 + 9;
bool isprime(long long n) {
  if (n <= 1) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return 1;
}
long long intlog(long long base, long long x) { return (log(x) / log(base)); }
bool prime[N];
void primee() {
  for (int p = 2; p * p <= N; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i < N; i += p) prime[i] = false;
    }
  }
}
string decToBinary(long long n) {
  string z = "";
  int i = 0;
  while (n > 0) {
    z += (n % 2) + 48;
    n = n / 2;
    i++;
  }
  reverse(((z).begin()), ((z).end()));
  return z;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, int y) {
  long long temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
int n, ans = oo;
vector<int> v;
void choose(int x) {
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += (2 * fabs(x - i) + 2 * (i - 1) + 2 * (x - 1)) * v[i];
  }
  ans = min(ans, sum);
}
int main() {
  scanf("%d", &n);
  v.resize(1 + n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
  }
  for (int i = 1; i <= n; i++) {
    choose(i);
  }
  printf("%d\n", ans);
  return 0;
}
