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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        double[][] a = new double[4][3];
        a[0][0] = 0.25;
        a[1][0] = 0.5;
        a[2][0] = 1.0;
        a[3][0] = 2.0;
        a[0][1] = sc.nextDouble();
        a[1][1] = sc.nextDouble();
        a[2][1] = sc.nextDouble();
        a[3][1] = sc.nextDouble();
        a[0][2] = a[0][1] / a[0][0];
        a[1][2] = a[1][1] / a[1][0];
        a[2][2] = a[2][1] / a[2][0];
        a[3][2] = a[3][1] / a[3][0];
        Arrays.sort(a, (x, y) -> Double.compare(x[2], y[2]));
        long n = sc.nextLong();
        long ans = 0;
        for (int i = 0; i < 4; i++) {
            ans += (long)a[i][1] * Math.floor(n/ a[i][0]);
            n -= (long)a[i][0] * Math.floor(n/ a[i][0]);
        }
        System.out.println(ans);
        sc.close();
    }

}