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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long q = sc.nextLong();
		long h = sc.nextLong();
		long s = sc.nextLong();
		long d = sc.nextLong();
		long[] price = { q * 4, h * 2, s, d / 2 };
		Arrays.sort(price);
		long n = sc.nextLong();
		long ans = 0;
		for (int i = 0; i < 4&&n!=0; i++) {
			if(price[i]==q*4) {
				ans += q*n*4;
				n -= n;
			}
			else if(price[i]==h*2) {
				ans += h*n*2;
				n -= n;
			}
			else if(price[i]==s) {
				ans += s*n;
				n -= n;
			}
			else if(price[i]==d/2) {
				ans += d*(n/2);
				n-=n/2*2;
			}
		}
		System.out.println(ans);
	}
}