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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;
typedef long long ll;
int main (void) {
    ll Q;
    ll H;
    ll S;
    ll D;
    ll N;

    cin >> Q >> H >> S >> D;
    cin >> N;

    double q = Q * 4;
    double h = H * 2;
    double s = S;
    double d = D / 2;
    double mi = min(min(min(q, h), s), d);

    ll ans = 0;
    if (mi == q) {
        ans += mi * N;
    } else if (mi == h) {
        ans += mi * N;
    } else if (mi == s) {
        ans += mi * N;
    } else {
        if (N >= 2) {
            ans += D * (N / 2);
            N -= N / 2 * 2;
            ans += min(min(q, h), s) * N;
        } else {
            ans += min(min(q, h), s) * N;
        }
    }

    cout << ans << endl;

    return 0;
}