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: Time Limit Exceeded)

#include<bits/stdc++.h>                                                            
using namespace std;

struct tset{
  int rate;
  int yen;
  double amount;
};

bool operator<(const tset& left, const tset& right){
  return left.rate < right.rate;
}

tset make(int r, int y, float a){
  tset s = {r, y, a};
  return s;
}

int main(){
  int q, h, s, d;
  double n;
  cin >> q >> h >> s >> d >> n;
  vector<tset> p(4);

  p[0] = make(q * 8, q, 0.25);
  p[1] = make(h * 4, h, 0.5 );
  p[2] = make(s * 2, s, 1.0 );
  p[3] = make(d    , d, 2.0 );

  sort(p.begin(), p.end());

  unsigned long long int sum = 0;
  for(int i = 0; i < 4; i++){
    while(p[i].amount <= n){
      n -= p[i].amount;
      sum += p[i].yen;
    }
  }

  cout << sum << endl;
  return 0;
}