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 <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define loop(i,m,n) for(int i=m;i<n;i++)
#define endre getchar(); getchar();return 0;
#define pie 3.141592653589793238462643383279
#define INF 1000000007
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
int lcm(int x, int y) {
	return x * y / gcd(x, y);
}
signed main() {
	int a, b, c, d,n;
	cin >> a >> b >> c >> d;
	cin >> n;
	a *= 4;
	b *= 2;
	int ans = min(a, b);
	ans = min(c, ans);
	if (n == 1) {
		cout << ans << endl;
	}
	else {
		if (d < ans * 2)cout << n / 2*d + ans << endl;
		else cout << ans*n << endl;
	}
}