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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

int compare_ll(const void *a, const void *b) {
  return *(long long*)a - *(long long*)b;
}

int main() {
  long long q, h, s, d, n;
  int flag = 0;
  scanf("%lld %lld %lld %lld", &q, &h, &s, &d);
  scanf("%lld", &n);
  long long pertwo[4];
  pertwo[0] = q*8;
  pertwo[1] = h*4;
  pertwo[2] = s*2;
  pertwo[3] = d;
  qsort(pertwo, 4, sizeof(long long), compare_ll);
  if (pertwo[0] == d) {
    if (pertwo[1] == d) {
      flag = 1;
    } else {
      flag = 2;
    }
  } else {
    flag = 1;
  }
  if (flag == 1) {
    long long ans = n*pertwo[0] / 2;
    printf("%lld", ans);
  }
  if (flag == 2) {
    if (n % 2 == 0) {
      long long ans = n*pertwo[0] / 2;
      printf("%lld", ans);
    } else {
      long long ans = (n-1)*pertwo[0] / 2 + pertwo[1]/2;
      printf("%lld", ans);
    }
  }
  return 0;
}