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>

int main(){
	int q,h,s,d;
	long long n,a,sum=0;
	
	scanf("%d %d %d %d",&q,&h,&s,&d);
	scanf("%ld",&n);
	
	a=d;
	if(a>s*2) a=s*2;
	if(a>h*4) a=h*4;
	if(a>q*8) a=q*8;
	
	while(n>=2){
		n-=2;
		sum+=a;
	}
	a=s;
	if(a>h*2) a=h*2;
	if(a>q*4) a=q*4;
	while(n>=1){
		n-=1;
		sum+=a;
	}
	a=h;
	if(a>q*2) a=q*2;
	while(n>=0.5){
		n-=0.5;
		sum+=a;
	}
	if(n!=0) sum+=q;
	printf("%ld",sum);

	return 0;
}