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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long[][] a = new long[4][3];
        a[0][0] = sc.nextLong();
        a[1][0] = sc.nextLong();
        a[2][0] = sc.nextLong();
        a[3][0] = sc.nextLong();
        a[0][1] = a[0][0] *8;
        a[1][1] = a[1][0] *4;
        a[2][1] = a[2][0] *2;
        a[3][1] = a[3][0];
        a[0][2] = 4;
        a[1][2] = 2;
        a[2][2] = 1;
        a[3][2] = 0;
        Arrays.sort(a, (x, y) -> Long.compare(x[1], y[1]));
        long n = sc.nextLong();
        long ans = 0;
        if(a[0][2] == 0){
            ans += Math.floorDiv(n, 2l) * a[0][0];
            if(n % 2 != 0){
                ans += a[1][0] * a[1][2];
            }
        }else{
            ans = n * a[0][0] * a[0][2];
        }
        System.out.println(ans);
        sc.close();
    }

}