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 <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
// *テンプレート*
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define P pair<int, int>
// aよりbの方が小さかったら、aの値をbの値に置き換える。(swapではない！)
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
// aよりbの方が大きかったら、aの値をbの値に置き換える。
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
const long long MOD = 1000000007;
const long long INF = 1LL << 29;
using namespace std;
typedef long long ll;
map<int, int> mp;
using Graph = vector<vector<int>>; 
//最大公約数
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}
//最小公倍数
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}

// *テンプレートここまで*

// pair -> vector<pair<int,int>> 
//		   AandB.push_back(make_pair(a[i],b[i]))


signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int q, h, s, d, n; cin >> q >> h >> s >> d >> n;
	cout << (n / 2)*min({ 8 * q,4 * h,2 * s,d }) + (n % 2)*min({ 4 * q,2 * h,s }) << endl;
	return 0;
}