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.

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 $n$ ($1 \leq n \leq 100$) — the number of floors.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 100$) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

见题面 Input 部分

见题面 Output 部分

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). $4 \cdot 2 + 8 \cdot 1 = 16$.

In the second example, the answer can be achieved by choosing the first floor as the $x$-th floor.