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 mpow(int base, long long exp);
void ipgraph(int m);
void DFS(int n);
bool isprime(int n);
int gcd(int a, int b);
void rstgraph(int n);
const int mod = pow(10, 9) + 7;
const int N = 3e5, M = N;
vector<int> g[N];
int vis[N];
int nodes = 0;
void iluzn() {
  int n;
  cin >> n;
  vector<int> a(n);
  int i;
  for (i = 0; i < n; i++) cin >> a[i];
  auto x = max_element(a.begin(), a.end());
  int d = x - a.begin();
  int moves = 0;
  for (i = 0; i < n; i++) {
    if (a[i] != 0) moves += ((i) + d + abs(d - i)) * a[i] * 2;
  }
  cout << moves << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t = 1, i;
  while (t--) iluzn();
  return 0;
}
int mpow(int base, long long exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
bool isprime(int n) {
  if (n < 2) return false;
  for (int x = 2; x * x <= n; x++) {
    if (n % x == 0) return false;
  }
  return true;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void ipgraph(int m) {
  int u, v;
  while (m--) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void DFS(int n) {
  ++nodes;
  vis[n] = 1;
  for (vector<int>::iterator i = g[n].begin(); i != g[n].end(); i++) {
    if (!vis[*i]) DFS(*i);
  }
}
void rstgraph(int n) {
  int i;
  for (i = 0; i < n + 1; i++) {
    g[i].clear();
    vis[i] = 0;
  }
  nodes = 0;
}
