Description: 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 Format: 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 Format: In a single line, print the answer to the problem — the minimum number of electricity units. 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). $$$4 \cdot 2 + 8 \cdot 1 = 16$$$. In the second example, the answer can be achieved by choosing the first floor as the $$$x$$$-th floor.