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 <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <queue>
#include <set>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
 
using namespace std;
typedef long long int lli;
const int MAX=1e5;
 
int main(){
	lli A,B,C,D,N;
	cin>>A>>B>>C>>D>>N;
  	N=4*N;
	if(2*A<B)B=2*A;
	if(2*B<C)C=2*B;
	if(2*C<D)D=2*C;
	lli a,b,c,d;
	d=N/8;
	c=(N%8)/4;
	b=(N%4)/2;
	a=N%2;
	lli ans=a*A+b*B+c*C+d*D;
	cout<<ans<<endl;
}