Codeforces 557B Pasha and Tea （思维）

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 ai 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 ai (1 ≤ ai ≤ 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 41 1 1 1

output

3

input

3 184 4 4 2 2 2

output

18

input

1 52 3

output

4.5

Note

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

题意：有相同数目为n的男生和 女生，和总数为2*n的茶杯（茶杯的容量不一定相等），和一个装着w茶水的茶壶.

若女生得到的茶水总数为x，则男生得到的茶水为2*x，问男生和女生能得到的茶水总数是多少.

思路：
先将各茶杯的容量存在一个数组中，然后从小到大排序，然后求出前n个每个茶杯都能得到最多茶水的数目，即在排序后的2*n个杯子中第一个杯子的容量a[0],和第n+1个杯子的容量的1/2  （a[n]/2）选出最小的.最后算出总数即可.

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[200200];int n,w;int main(){while(scanf("%d%d",&n,&w)!=EOF){int i;for(i=0;i<n*2;i++){scanf("%d",&a[i]);}sort(a,a+n*2);double gv,bv,cv;gv=min(a[0]*1.0,a[n]*1.0/2);bv=gv*2;cv=(gv+bv)*n;cv=min(cv,w*1.0);printf("%.6lf\n",cv);}}