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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  ll Q, H, S, D, N;
  cin >> Q >> H >> S >> D >> N;
  ll one = min(min(4 * Q, 2 * H), S);
  ll two = min(2 * one, D);
  cout << (N / 2) * two + (N % 2) * one << endl;
  return 0;
}