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: Accepted)

import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int q = sc.nextInt();
        int h = sc.nextInt();
        int s = sc.nextInt();
        int d = sc.nextInt();
        int n = sc.nextInt();
        int m;
        if (q * 2 < h) {
            m = q * 4;
        } else {
            m = h * 2;
        }
        if (m > s) {
            m = s;
        }
        if (d > m * 2) {
            d = m * 2;
        }
        BigDecimal ans = new BigDecimal("0");
        BigDecimal two = new BigDecimal("2");
        BigDecimal bn = BigDecimal.valueOf(n);
        BigDecimal hn = BigDecimal.valueOf(n / 2);
        BigDecimal bm = BigDecimal.valueOf(m);
        BigDecimal bd = BigDecimal.valueOf(d);
        if (n % 2 == 0) {
            ans = hn.multiply(bd);
        } else {
            ans = hn.multiply(bd);
            ans = ans.add(bm);
        }
        System.out.println(ans);
    }
}