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: Accepted)

#include <bits/stdc++.h>
#define RG register
#define Min(a,b) (a<b?a:b)
#define ll long long

using namespace std;

inline int gi(){
	RG char ch=getchar();RG int x=0,q=0;
	while(ch<'0' || ch>'9') ch=='-'?(q=1):0,ch=getchar();
	while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
	return q?(-x):x;
}

int main(){
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	int q=gi(),h=gi(),s=gi(),d=gi(),n=gi();
	h=Min(h,(q<<1));
	s=Min(s,(h<<1));
	d=Min(d,(s<<1));
	ll ans=1ll*n/2*d;
	if (n&1) ans+=s;
	cout<<ans;
	return 0;
}