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)

#define _CRT_SECURE_NO_DEPRECATE
#define rep(i, b, e) for (auto i = b; i < e; ++i)
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#ifndef __GNUC__
#include <intrin.h>
#endif // !__GNUC__
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <numeric>
#include <cstring>
#include <complex>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <functional>

int bsf(int x) {
#ifdef __GNUC__
    return __builtin_ctz(x);
#else
    unsigned long i;
    _BitScanForward(&i, x);
    return i;
#endif
}
int bsr(int x) {
#ifdef __GNUC__
    return 31 - __builtin_clz(x);
#else
    unsigned long i;
    _BitScanReverse(&i, x);
    return i;
#endif
}

using namespace std;
typedef long long int64;
const double pi = 3.14159265358979323846;
const int inf = (int)2e+9 + 2;
const int mod = (int)1e+9;
const double eps = 1e-8;
const int N = 1 << 20;

int n, m, k;
int c[4];

void solve()
{
    rep(i, 0, 4)
        cin >> c[i];
    cin >> n;
    int best1 = c[2], best2 = c[3];
    best1 = min(best1, c[0] * 4);
    best1 = min(best1, c[1] * 2);
    best2 = min(best2, best1 * 2);
    if (n & 1)
        cout << (int64)(n / 2) * best2 + best1;
    else
        cout << (int64)(n / 2) * best2;
}

int main()
{
#ifndef ONLINE_JUDGE
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // !ONLINE_JUDGE
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int test = 1;
    //cin >> test;
    for (; test; --test)
        solve();
    return 0;
}