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 <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <bitset>
#include <sstream>
#include <cstdlib>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <climits>
#include <ctype.h>
using namespace std;

#define PI acos(-1.0)
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define fi first
#define se second

#define eps 1e-8
#define mod 1000000007

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<vector<int>> vvi;

struct Matrix {
    vvi data;
    int r, c;
    Matrix(int row, int col, bool identity = false) : r(row), c(col) {
        data.assign(row, vector<int>(col, 0));
        if (identity) {
            for (int i = 0; i < r; ++i) {
                data[i][i] = 1;
            }
        }
    }
    Matrix operator * (Matrix& other) {
        int m = r, n = c, p = other.c;
        Matrix res(m, p);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < p; ++j) {
                for (int k = 0; k < n; ++k) {
                    res.data[i][j] += data[i][k] * other.data[k][j];
                }
            }
        }
        return res;
    }
};

// 扩展欧几里得算法 res[1] * a + res[2] * b = res[0] = gcd(a, b)
vector<int> extendGcd(int a, int b) {
    if (b == 0) {
        return {a, 1, 0};
    } else {
        vector<int> tmp = extendGcd(b, a % b);
        return {tmp[0], tmp[2], tmp[1] - (a / b) * tmp[2]};
    }
}

// 矩阵快速幂
Matrix matrix_power(Matrix base, ll exp) {
    int n = base.r;
    Matrix res(n, n, true);
    while (exp) {
        if (exp & 1) {
            res = res * base;
        }
        base = base * base;
        exp >>= 1;
    }
    return res;
}

// 带模快速幂
ll power_mod(ll base, int exp) {
    ll res = 1;
    while (exp) {
        if (exp & 1) res = res * base % mod;
        base = base * base % mod;
        exp >>= 1;
    }
    return res;
}

// 快速幂
ll power(ll base, int exp) {
    ll res = 1;
    while (exp) {
        if (exp & 1) res *= base;
        base *= base;
        exp >>= 1;
    }
    return res;
}

// 求逆元
ll inv(ll a) {
    return power_mod(a, mod - 2);
}

#define MAXFAC 100010

ll fac[MAXFAC];

void initFac() {
    fac[0] = 1;
    for (int i = 1; i < MAXFAC; ++i) {
        fac[i] = i * fac[i - 1] % mod;
    }
}

ll Combine(ll a, ll b) {
    return (fac[a] * inv(fac[b]) % mod) * inv(fac[a - b]) % mod;
}

/******************************** template ********************************/
#define MAXN 2010
int q, h, s, d;
int n;

int main() {
    cin >> q >> h >> s >> d;
    cin >> n;
    int tmp = min(4 * q, min(2 * h, s));

    if (d < 2 * tmp) {
        ll res = 1LL * (n / 2) * d;
        if (n % 2) res += tmp;
        cout << res;
    } else {
        cout << 1LL * n * tmp;
    }

    return 0;
}