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 long long int N = (long long int)2e5 + 10;
long long int solve(istream &in, ostream &out) {
  multiset<long long int> s;
  long long int n, val = 0, cap = 0, sum = 0;
  in >> n >> cap;
  long long int a[N] = {};
  for (long long int i = 0; i < n; i++) in >> a[i], sum += a[i], s.insert(a[i]);
  if (sum < cap) return puts("-1");
  long long int aa = *s.begin();
  auto iter = s.end();
  iter--;
  long long int bb = *iter;
  while (bb - aa != 0 and cap != 0) {
    if (cap >= bb - aa)
      cap -= (bb - aa);
    else {
      cap = 0;
      iter = s.end();
      iter--;
      s.erase(iter);
      s.insert(bb - cap);
      break;
    }
    iter = s.end();
    iter--;
    s.erase(iter);
    s.insert(aa);
    aa = *s.begin();
    iter = s.end();
    iter--;
    bb = *iter;
  }
  val = *s.begin();
  if (cap == 0) {
    out << val << "\n";
    return 0;
  }
  long long int no_of = cap / n;
  if (no_of == 0)
    out << val - 1 << "\n";
  else
    out << val - no_of << "\n";
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(16);
  solve(cin, cout);
}
