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.io.*;
import java.util.*;
import java.math.*;
public class Main {
  public static void main(String[] args) throws Exception {
    String line = null;
    double num = 0;
    double rem = 0;
    double res = 0;
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    line = br.readLine();
    String price[] = line.split(" ");
    line = br.readLine();
    long vol = Long.parseLong(line);
    // 配列に1リットルあたりの単価を登録
    long cost1 = Long.parseLong(price[0]) * 4;         // 0.25L
    long cost2 = Long.parseLong(price[1]) * 2;         // 0.50L
    long cost3 = Long.parseLong(price[2]);             // 1.00L
    long d_cost4 = Long.parseLong(price[3]);           // 2.00L * 2
    // 0.25L, 0.5L. 1Lの単価の最小値
    long minPrice = Math.min(Math.min(cost1, cost2), cost3);
    if (vol < 2) {
      res = minPrice * vol;
    } else {
      if (minPrice * 2 < d_cost4) {
        res = minPrice * vol;
      } else {
        num = vol / 2;  // 2Lで買える数
        rem = vol % 2;  // 2Lで買えない数(1 or 0)
        res = (d_cost4 * num) + (minPrice * rem);
      }
    }
    System.out.println((long)res);
  }

}