Codeforces Round #382 (Div. 2) 735B Urbanization

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B. Urbanization

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 aren people who plan to move to the cities. The wealth of thei of them is equal to a_i. Authorities plan to build two cities, first forn₁ people and second forn₂ people. Of course, each ofn candidates can settle in only one of the cities. Thus, first some subset of candidates of sizen₁ settle in the first city and then some subset of sizen₂ 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 thatthe 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 wealtha_i among all its residents divided by the number of them (n₁ orn₂ 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.

Input

The first line of the input contains three integers n,n₁ and n₂ (1 ≤ n, n₁, n₂ ≤ 100 000,n₁ + n₂ ≤ 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 a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100 000), thei-th of them is equal to the wealth of thei-th candidate.

Output

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 exceed10^- 6.

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

Examples

Input

2 1 11 5

Output

6.00000000

Input

4 2 11 4 2 3

Output

6.50000000

Note

In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.

In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (a₃ + a₄) / 2 + a₂ = (3 + 2) / 2 + 4 = 6.5

题意：有一个人要规划住房。总共有n个人。每个人都有财富值。规划者要把这n个人规划到两个城市里。第一个城市能住n1个人，第二个城市能住n2个人。如果人城市住满了。那么剩下的人会被遣返回原来的住处。要求是让这个两个城市的财富平均值总和尽量大。一个城市的财富平均值为：(城市人口财富总和)/（总人数）

思路：先从小到大排个序：把财富值大的几个人安到人少的那个城市，财富值相对小的安到人较多的城市。

#include<bits/stdc++.h>using namespace std;int main(){    int n1,n2,n;    double wel[100005];    while(~scanf("%d %d %d",&n,&n1,&n2))    {        for(int i=0; i<n; i++)        {            scanf("%lf",&wel[i]);        }        sort(wel,wel+n);        int MAX=max(n1,n2);        int MIN=min(n1,n2);        double sum1=0;        double sum2=0;        for(int i=0; i<MIN; i++)        {            sum1+=wel[n-1-i];        }        for(int i=0; i<MAX; i++)        {            sum2+=wel[n-MIN-1-i];        }        printf("%.8lf\n",(sum1/(MIN*1.0))+(sum2/(MAX*1.0)));    }    return 0;}