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 (Clang 3.8.0) (judge verdict: Accepted)

#include<cstdio>
#define int long long
int a,b,c,d,n;
signed main()
{
	scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&n);
	int w=a*8,x=b*4,y=c*2,z=d;
	if(z<=w&&z<=x&&z<=y)
	{
		int ans=n>>1;n&=1;
		if(!n)printf("%lld\n",ans*d);
		else if(w<=x&&w<=y)printf("%lld\n",ans*d+a*4);
		else if(x<=w&&x<=y)printf("%lld\n",ans*d+b*2);
		else printf("%lld\n",ans*d+c);
	}
	else if(w<=x&&w<=y&&w<=z)printf("%lld\n",a*4*n);
	else if(x<=w&&x<=y&&x<=z)printf("%lld\n",b*2*n);
	else printf("%lld\n",c*n);
	return 0;
}