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 <iostream>
#include <ccomplex>
#include <vector>
#include <cassert>
#include <utility>
#include <algorithm>
#include <string>
#include <deque>
#include <queue>
#include <functional>
#include <cmath>
#include <iomanip>
#include <map>
#include <numeric>
#include <list>
#include <assert.h>
#include <math.h>
#include <valarray>
#include <stdio.h>
#include <algorithm>
#include <set>
#include <complex>
#include <list>
#include <time.h>
#include <stack>
#include <locale>
#include <clocale>
#include <ctype.h>
#include <wchar.h>

using namespace std;
typedef long long int LL;
typedef pair<long long int, long long int> pii;
typedef pair<double, double> pdd;

#define SORT(c) sort((c).begin(),(c).end())
#define BACKSORT(c) sort((c).begin(),(c).end(),std::greater<LL>())
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)


int xcomp(vector<LL>& a, vector<LL>& b) {
	return a[0] < b[0];
}

#define XSORT(c) sort((c).begin(),(c).end(),xcomp)

#define ALL(x) (x).begin(),(x).end()
LL mod = 1000000007;

int main() {
	LL Q, H, S, D, N;
	cin >> Q >> H >> S >> D >> N;
	N *= 4;

	LL sum = 0;

	if (N >= 16) {
		LL mini = min({ Q * 8, H * 4, S * 2, D });
		LL kosuu = ((N - 8) / 8);
		sum += mini*kosuu;
		N -= kosuu * 8;
	}


	LL mini2 = 1000000000000;

	REP(i,17){
		REP(j, 17) {
			REP(k, 17) {
				REP(l, 17) {
					if (i + j * 2 + k * 4 + l * 8 == N) {
						if (i*Q + j*H + k*S + l*D < mini2) {
							mini2 = i*Q + j*H + k*S + l*D;
						}
					}
				}
			}
		}
	}
	sum += mini2;

	cout << sum << endl;
}