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>

using namespace std;
typedef long long int ll;
typedef vector<int> vi;
typedef vector<ll> vl;

const int INF = 1e9;
const ll lINF = 1e18;
const int MOD = 1e9+7;
const int SIZE = 100;


int main() {

	vector<pair<ll, int> > p;
	ll a[4];
	ll ans = 0;
	for (int i = 0; i < 4; i++) {
		cin >> a[i];
		p.push_back(make_pair(a[i], i));
	}
	ll n;
	cin >> n;
	for (int i = 0; i < 4; i++) {
		p[i].first *= (1 << (3 - i));
	}
	sort(p.begin(), p.end());

	for (int i = 0; i < 4; i++) {
		p[i].first /= (1 << (3 - p[i].second));
	}

	if (p[0].second == 3) {
		ans += (n / 2) * p[0].first;
		n %= 2;
		ans += n * p[1].first * (1 << (2 - p[1].second)); 
	} else {
		ans += n * p[0].first * (1 << (2 - p[0].second)); 
	}

	cout << ans << endl;
}