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: Accepted)

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
using namespace std;

int main(){

    long long q, h, s, d; cin >> q >> h >> s >> d;

    int n; cin >> n;

    long long a_2l[] = {8*q, 4*h, 2*s, d};

    long long minv = 1e10;
    int i_2l = 0;
    int i_1l = 0;
    for(int i=0; i<4; i++){
        if(minv > a_2l[i]){
            minv = a_2l[i];
            i_2l = i;
            if(i < 3){
                i_1l = i;
            }
        }
    }

    long long ans = (n / 2) * a_2l[i_2l];
    if(n % 2 == 1){
        ans += a_2l[i_1l]/2;
    }
       
    cout << ans << endl;

    return 0;
}