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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, int> pli;
typedef pair <int, ll> pil;
typedef pair <ll, ll> pll;
typedef pair <int, pii> piii;
#define fr first
#define sc second
#define pb push_back
#define ppb pop_back()
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define sz(x) int(x.size())
#define mpr make_pair
ll a[4], n;
pair <double, ll> b[4];
int main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	ios_base :: sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	for(int i = 0;i < 4; ++ i){
		cin >> a[i];
		b[i].fr = (1.0 * a[i]) / (1.0 * (1 << i));
		b[i].sc = i;
	}
	cin >> n;
	n *= 4;
	sort(b, b + 4);
	if(b[0].sc < 3 || (n / 4) % 2 == 0){
		cout << n * b[0].fr;
		return 0;
	}
	cout << (n - 4) * b[0].fr + 4 * b[1].fr;
	return 0;
}