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>
//#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>
#include <numeric>
#include <iomanip>
 
#define FOR(i, a, b) for( int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for( int i=(a);i>=(b);i--)
#define LFOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define LRFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
using namespace std;
using ll=long long;
constexpr ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
constexpr ll lcm(ll a,ll b){return a*b/gcd(a,b);}
 
int main() {
ll Q,H,S,D;
ll N;

ll ans=0;
  cin >> Q>>H>>S>>D;
  cin >> N;

ll saiyasu;

 saiyasu =Q*4;
 if(Q*4>H*2){ saiyasu =H*2;};
 if(saiyasu>S){ saiyasu =S;};

ans= (N/2)*D+(N%2)*saiyasu;

if(ans>N*saiyasu){ans=N*saiyasu;}


  cout << ans<< endl;
  return 0;
}