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.*;
import java.lang.*;

import static java.lang.StrictMath.abs;

class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long q = sc.nextInt();
        long h = sc.nextInt();
        long s = sc.nextInt();
        long d = sc.nextInt();
        long n = sc.nextInt();
        long t = Math.min(8*q, 4*h);
        long t2 = Math.min(t, 2*s);
        long t3 = Math.min(t2, d);
        long ans = 0;
        if(t3 == 8*q){
            ans = q*4*n;
        }else if(t3 == 4*h){
            ans = h * 2 * n;
        }else if(t3 == s*2){
            ans = s*n;
        }else if(t3 == d){
            ans += d*(n/2);
            if(t2 == 2*s){
                ans += s;
            }else if(t2 == 4*h){
                ans += 2*h;
            }else if(t2 == 8*q){
                ans += 4*q;
            }
        }
        System.out.println(ans);

        


    }
}