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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<utility>
#include<numeric>
#include<vector>
#include<map>
#include<unordered_map>
#include<set>
#include<tuple>
#include<stack>
#include<queue>
#include<tuple>
#include<functional>
#include<iterator>
#include<cmath>
#include<cctype>

using namespace std;

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

const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = INF+7;

struct edge{int to,cost;};

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	int q,h,s,d,n;
	cin >> q >> h >> s >> d >> n;

//	pair<double,double> vl_l_v[4];
	tuple<double,double,int> vl_l_v[4];
	vl_l_v[0]=make_tuple(8.0*q,0.25,q);
	vl_l_v[1]=make_tuple(4.0*h,0.5,h);
	vl_l_v[2]=make_tuple(2.0*s,1,s);
	vl_l_v[3]=make_tuple(d*1.0,2,d);

	sort(vl_l_v,vl_l_v+4);

	ll ans=0;
	int idx=0;
	double vl,l;
	ll v;

	tie(vl,l,v) = vl_l_v[0];
	if(l == 2.0 && n%2 == 1){
		ans += n/2*d;
		n=1;
		idx++;
//		cout << "2L " << ans << "\n";
	}
	tie(vl,l,v) = vl_l_v[idx];
	ans += n/l * v;
	cout << ans << "\n";
		
	return 0;
}