Codeforces 557B Pasha and Tea【二分+思维】

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本文介绍了一种解决茶话会上水量公平分配的算法。给定不同容量的茶杯及总量，目标是向男女宾客公平分配茶水，使得每位女士得到的茶水量为男士的一半，并最大化总分配量。

Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of w milliliters and 2n tea cups, each cup is for one of Pasha's friends. The i-th cup can hold at most a_i milliliters of water.

It turned out that among Pasha's friends there are exactly n boys and exactly n girls and all of them are going to come to the tea party. To please everyone, Pasha decided to pour the water for the tea as follows:

Pasha can boil the teapot exactly once by pouring there at most w milliliters of water;
Pasha pours the same amount of water to each girl;
Pasha pours the same amount of water to each boy;
if each girl gets x milliliters of water, then each boy gets 2x milliliters of water.

In the other words, each boy should get two times more water than each girl does.

Pasha is very kind and polite, so he wants to maximize the total amount of the water that he pours to his friends. Your task is to help him and determine the optimum distribution of cups between Pasha's friends.

Input

The first line of the input contains two integers, n and w (1 ≤ n ≤ 10⁵, 1 ≤ w ≤ 10⁹) — the number of Pasha's friends that are boys (equal to the number of Pasha's friends that are girls) and the capacity of Pasha's teapot in milliliters.

The second line of the input contains the sequence of integers a_i (1 ≤ a_i ≤ 10⁹, 1 ≤ i ≤ 2n) — the capacities of Pasha's tea cups in milliliters.

Output

Print a single real number — the maximum total amount of water in milliliters that Pasha can pour to his friends without violating the given conditions. Your answer will be considered correct if its absolute or relative error doesn't exceed 10^- 6.

Examples

Input

2 4
1 1 1 1

Output

Input

3 18
4 4 4 2 2 2

Output

Input

1 5
2 3

Output

4.5

Note

Pasha also has candies that he is going to give to girls but that is another task...

题目大意：

给你N个杯子，每个杯子的最大容量已知，共有WmL的tee，想分给n个女孩和n个男孩，要求：

①给每个女孩的量都要相同。

②给每个男孩的量都要相同。

③给每个女孩的量是每个男孩的量的1/2.

问最多可以分配出去多少mL的tee.

思路：

1、首先我们按照杯子的容量从小到大排序.那么前n个杯子我们给女生用，后n个杯子给男生用。因为男生的量要大于女孩的量。

2、那么接下来考虑这样一点，对于n个女孩子的杯子，很显然最多每个女孩能够分到a【0】容量的tee.每个男孩最多能够分到a【n】容量的tee.

我们可以通过枚举给每个女孩的量来判断是否可行，直到不可行为止，那么这个值很显然就是最多的量。

那么分析出，对于给每个女孩的量越大，越有可能不能分配成功，那么我们可以二分这个值。

对于当前二分出来的值mid.表示给每个女孩的量，那么需要满足：mid*n*3<=w,mid<=a[0]&&2*mid<=a[n];

如果当前值可行，那么加大量，否则减少量。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define eps 1e-11
double a[300050];
int main()
{
    int n;
    double w;
    while(~scanf("%d%lf",&n,&w))
    {
        for(int i=0;i<n*2;i++)scanf("%lf",&a[i]);
        sort(a,a+n*2);
        double ans=-1;
        double l=0;
        double r=1000000000;
        while(r-l>=eps)
        {
            double mid=(l+r)/2;
            if(mid*n*3<=w+eps&&a[0]+eps>=mid&&a[n]+eps>=2*mid)
            {
                ans=mid;
                l=mid;
            }
            else r=mid;
        }
        printf("%lf\n",ans*n*3);
    }
}