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
{
    long q, h, s, d;
    long n;
    public static void main(String[] args) {
        Main m = new Main();
        m.read();
        m.solve();
    }

    private void read() {
        Scanner sc = new Scanner(System.in);
        q = sc.nextLong();
        h = sc.nextLong();
        s = sc.nextLong();
        d = sc.nextLong();
        n = sc.nextLong();
    }

    private void solve() {
        n *= 4;
        Pair[] ps = new Pair[4];
        ps[0] = new Pair(q, 1);
        ps[1] = new Pair(h, 2);
        ps[2] = new Pair(s, 4);
        ps[3] = new Pair(d, 8);
        Arrays.sort(ps);
        long ans = 0L;
        for (int i = 0; i < 4 && n > 0; i++) {
            long tmp = n / ps[i].snd;
            n %= ps[i].snd;
            ans += (tmp * ps[i].fst);
        }
        System.out.println(ans);
    }

    private class Pair implements Comparable<Pair>{
        long fst, snd;
        Pair(long f, long s) {
            fst = f;
            snd = s;
        }

        @Override
        public int compareTo(Pair pair) {
            return Double.compare((double)fst / snd, (double)pair.fst/pair.snd);
        }
    }
}