Codeforces#382(Div. 2) B.Urbanization【贪心】

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探讨如何将N个人根据其财富值最优分配至两个城市，使两城居民平均财富之和最大化的问题及解决方案。

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 n₁ people and second for n₂ people. Of course, each of n candidates can settle in only one of the cities. Thus, first some subset of candidates of size n₁ settle in the first city and then some subset of size n₂ 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 (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 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), the i-th of them is equal to the wealth of the i-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 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 $\text{[math]}$ .

Examples

Input

2 1 1
1 5

Output

6.00000000

Input

4 2 1
1 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

题目大意：

一共有N个人，现在其中需要将其分配到两个城市中去，每个人都有其权值，城市1需要n1个人，城市2需要n2个人，对应每个城市的权值为其平均值，问总和的权值最大是多少。

思路：

1、首先我们按照每个人的权值排序，接下来思考：

①显然平均值的分母越小，其城市的总值越大。那么我们取n1和n2中最小值记做minn，分配给其minn个人（取权值最大的minn个人），接下来再取n1和n2中最大值记做maxn，分配给其maxn个人（去掉那minn个人之后中最大的maxn个人）；

②贪心过程结束。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int a[100500];
int main()
{
    int n,n1,n2;
    while(~scanf("%d%d%d",&n,&n1,&n2))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n);
        int minn=min(n1,n2);
        int maxn=max(n1,n2);
        double ans=0;
        double ans2=0;
        int cont=0;
        int pos;
        for(int i=n-1;i>=0;i--)
        {
            cont++;
            ans+=a[i];
            if(cont==minn)
            {
                pos=i-1;
                break;
            }
        }
        cont=0;
        for(int i=pos;i>=0;i--)
        {
            cont++;
            ans2+=a[i];
            if(cont==maxn)
            {
                break;
            }
        }
        printf("%lf\n",ans/minn+ans2/maxn);
    }
}