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>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define rep(i,n) for(int i=0;i<=(n);++i)
#define ALL(v) (v).begin(),(v).end()

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

ll q,h,s,d,n;
priority_queue<P,vector<P>,greater<P> > pq;

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

	cin>>q>>h>>s>>d>>n;
	if(n==1) {
		ll mi=min(q*4,min(h*2,s));
		cout<<mi<<endl;
		return 0;
	}
	pq.push(P(q*8,1)); pq.push(P(h*4,1)); pq.push(P(s*2,1)); pq.push(P(d,2));
	ll two=n/2;
	ll one=n%2;
	if(one==0) {
		P p=pq.top();
		cout<<n/2*p.first<<endl;
		return 0;
	}
	else {
		ll ans=0;
		P p=pq.top();pq.pop();
		if(p.second==2) {
			ans+=n/2*p.first;
			P q=pq.top();
			ans+=q.first/2;
		}
		else {
			ans+=n*p.first/2;
		}
		cout<<ans<<endl;
		return 0;
	}

	return 0;
}