codeforces 735 B Urbanization

题目描述：

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 or n₂ 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 andn₂ (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.

Example

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

题目分析：排2次序就ok啦

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int n,n1,n2;int v[200005];int main(){    while(~scanf("%d%d%d",&n,&n1,&n2))    {        if(n1>n2)        {            int t=n1;            n1=n2;            n2=t;        }        for(int i=0;i<n;i++)            scanf("%d",&v[i]);        sort(v,v+n);        double sum1=0,sum2=0;        double p;        for(int i=n-1;i>=n-n1;i--)            sum1+=v[i];        for(int i=n-n1-1;i>=n-n1-n2;i--)            sum2+=v[i];        p=sum1/n1+sum2/n2;        //cout<<p<<endl;        printf("%lf\n",p);    }    return 0;}