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;
long long invm(long long a, long long m) {
  a = a % m;
  for (long long x = 1; x < m; x++)
    if ((a * x) % m == 1) return x;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long gcdv(vector<long long> v) {
  long long a = 0;
  vector<long long>::iterator it = v.begin();
  a = *it;
  it++;
  for (; it != v.end(); it++) {
    a = gcd(a, *it);
  }
  return a;
}
vector<long long> SOE(long long n) {
  vector<bool> prime;
  for (long long i = 0; i <= n; i++) prime.push_back(true);
  vector<long long> v;
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) v.push_back(p);
  return v;
}
vector<long long> primeFactors(long long n) {
  long long k = 0, f = 0;
  vector<long long> v;
  while (n > 1) {
    f = 0;
    vector<long long> tv(SOE(sqrt(n)));
    for (vector<long long>::reverse_iterator it = tv.rbegin(); it != tv.rend();
         it++) {
      if ((n % (*it)) == 0) {
        v.push_back(*it);
        n /= *it;
        f = 1;
        break;
      }
    }
    if (f == 0) {
      v.push_back(n);
      break;
    }
  }
  return v;
}
long long pow2(long long exp) {
  if (exp == 1) return 2;
  if (exp == 0) return 1;
  long long x = pow2(exp / 2);
  if (exp % 2 == 0) {
    return ((x % 1000001) * (x % 1000001)) % 1000001;
  } else {
    return ((x % 1000001) * (x % 1000001) * 2) % 1000001;
  }
}
int main() {
  int n;
  cin >> n;
  int a[n];
  int ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ans += (i * 4) * a[i];
  }
  cout << ans << endl;
}
