Problem:
Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            var a = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
            long n = long.Parse(Console.ReadLine());

            double q = 4.0 * a[0];
            double h = 2.0 * a[1];
            double s = 1.0 * a[2];
            double d = 0.5 * a[3];

            var dic = new Dictionary<double, double>();
            dic.Add(q, 0.25);
            dic.Add(h, 0.5);
            dic.Add(s, 1.0);
            dic.Add(d, 2.0);

            dic = dic.OrderBy(x => x.Key).ToDictionary(x => x.Key, x => x.Value);
            long m = 0;
            long ans = 0;
            for (int i = 0; i < 4; i++)
            {
                m = (long)(n / dic.Values.ElementAt(i));
                Console.WriteLine("{0}:{1}:{2}", m, dic.Values.ElementAt(i), n);
                ans += (long)(m * (dic.Keys.ElementAt(i) + 0.0000000000001) * dic.Values.ElementAt(i));
                n -= (long)(m * dic.Values.ElementAt(i));
            }
            Console.WriteLine(ans);
        }
    }
}