A10763 | Urbanization

Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are $n$ people who plan to move to the cities. The wealth of the $i$ of them is equal to $a_{i}$ . Authorities plan to build two cities, first for $n_{1}$ people and second for $n_{2}$ people. Of course, each of $n$ candidates can settle in only one of the cities. Thus, first some subset of candidates of size $n_{1}$ settle in the first city and then some subset of size $n_{2}$ is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.

To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth $a_{i}$ among all its residents divided by the number of them ( $n_{1}$ or $n_{2}$ depending on the city). The division should be done in real numbers without any rounding.

Please, help authorities find the optimal way to pick residents for two cities.

The first line of the input contains three integers $n$ , $n_{1}$ and $n_{2}$ ( $1<=n,n_{1},n_{2}<=100000$ , $n_{1}+n_{2}<=n$ ) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.

The second line contains $n$ integers $a_{1},a_{2},...,a_{n}$ ( $1<=a_{i}<=100000$ ), the $i$ -th of them is equal to the wealth of the $i$ -th candidate.

Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed $10^{-6}$ .

Namely: let's assume that your answer is $a$ , and the answer of the jury is $b$ . The checker program will consider your answer correct, if .