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 TEST_MODE true

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, a, b) for (int i = (a); i < (int)(b); ++i)
#define rrep(i, n) for (int i = (n)-1; i >= 0; --i)
#define rrep2(i, a, b) for (int i = (a)-1; i >= (int)b; --i)
#define range(i, a, b, c) for (int i = a;             \
                               c > 0 ? i < b : i > b; \
                               i += c)
#define chmax(a, b) (a = (a) < (b) ? (b) : (a))
#define chmin(a, b) (a = (a) > (b) ? (b) : (a))
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define all(a) begin(a), end(a)
#define ifnot(a) if (not(a))
#define int long long

#ifdef LOCAL_ENV

#if TEST_MODE == true
const bool test = true;
#define dump(x) cerr << #x << " = " << (x) << " "
#else
const bool test = false;
#define dump(x) 
#endif

#else
const bool test = false;
#define dump(x) 
#endif

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int INF = (int)1 << 60;
const ll INFL = (ll)1 << 60;
ll mod_n = (int)1e9 + 7;
const double eps = 1e-10;
typedef long double Real;
// return -1, 0, 1
int sgn(const Real &r) { return (r > eps) - (r < -eps); }
int sgn(const Real &a, const Real &b) { return sgn(a - b); }

//.....................
const int MAX = (int)2e5 + 5;

vector<string> split(const string &str, char sep)
{
    vector<string> v;
    stringstream ss(str);
    string buffer;
    while (getline(ss, buffer, sep))
    {
        v.push_back(buffer);
    }
    return v;
}

template <class InputIterator>
int sum(InputIterator begin, InputIterator end)
{
    return accumulate(begin, end, 0ll);
}

template<typename T>
T gcd(T a, T b)
{
	T c;
	while (a != 0) {
		c = a; a = b%a;  b = c;
	}
	return b;
}

void solve();

class Solver {
public:
    int Q, H, S, D;
    int N;
    int N2;

    void solve() {
        cin >> Q >> H >> S >> D;
        cin >> N;
        chmin(H, Q*2);
        chmin(S, H*2);
        chmin(D, S*2);
        int ans = D * (N / 2) + S * (N % 2);        
        cout << ans << endl;
    }
};

signed main()
{
    srand(time(NULL));
    cout << fixed << setprecision(20);
    auto ptr = new Solver();
    ptr->solve();
    delete ptr;
    // while (true) {    
    //     auto ptr = new Solver();
    //     ptr->solve();
    //     delete ptr;
    // }
    return 0;
}

// class Mycin {
// 	bool flag = true;
// public:
// 	Mycin& operator >> (int& a) {flag = scanf("%lld", &a) != EOF; return *this;}
// 	Mycin& operator >> (char& a) {flag = scanf("%c", &a) != EOF; return *this;}
// 	Mycin& operator >> (string& s) {flag = (bool)(cin >> s); return *this;}
// 	operator bool() {return flag;}
// } mycin;
 
// class Mycout {
// public:
// 	Mycout& operator << (const int& a) {printf("%lld", a); return *this;}
// 	Mycout& operator << (const char c) {printf("%c", c); return *this;}
// 	Mycout& operator << (const string& s) {printf("%s", s.c_str()); return *this;}
// } mycout;
 
// #define cin mycin
// #define cout mycout
// #define endl '\n'