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<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<string>
#include<queue>
#include<map>

using namespace std;

typedef long long ll;
typedef pair<int,int> P;

ll q,h,s,d,n;

ll tmin(ll a,ll b,ll c)
{
	ll k=min(a,b);
	return min(k,c);
}
ll fmin(ll a,ll b,ll c,ll d)
{
	ll k=tmin(a,b,c);
	return min(k,d);
}
int main()
{
	cin>>q>>h>>s>>d;
	cin>>n;
	int k=n/2;
	int l=n%2;

	ll ans=0;
	ans+=k*fmin(8*q,4*h,2*s,d);
	ans+=l*tmin(4*q,2*h,s);

	cout<<ans<<endl;

	return 0;
}