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>
#include <time.h>
#include <limits.h>
#include <stdint.h>

unsigned long long int price[4];
unsigned long long int n;

int main() {
	for(int i = 0; i < 4; i++) {
		scanf("%llu", &price[i]);
	}
	scanf("%llu", &n);
	if(n == 1) {
		price[0] *= 4;
		price[1] *= 2;

		unsigned long long int min = 1844674407370955161;
		for (int i = 0; i < 3; i++) {
			if(min > price[i]) {
				min = price[i];
			}
		}
		printf("%llu\n", min);
	}
	else {
		price[0] *= 8;
		price[1] *= 4;
		price[2] *= 2;

		unsigned long long int min = 1844674407370955161;
		for(int i = 0; i < 4; i++) {
			if(min > price[i]) {
				min = price[i];
			}
		}

		if(n % 2 == 0) {
			//余り0
			printf("%llu\n", n / 2 * min);
		}
		else {
			unsigned long long int min_new = 1844674407370955161;
			for (int i = 0; i < 3; i++) {
				if(min_new > price[i]) {
					min_new = price[i];
				}
			}
			printf("%llu\n", n / 2 * min + min_new / 2);
		}
	}
	return 0;
}