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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php

// define('DEBUG', true);
define('DEBUG', false);

fscanf(STDIN, "%d %d %d %d", $Q, $H, $S, $D);
fscanf(STDIN, "%d", $N);

$lit1 = min($Q * 4, $H * 2, $S);

if ($lit1 < $D / 2) {
    // $lit1ベースで買った方が安い
    ans($N * $lit1);
} else {
    if ($N % 2 == 0) {
        ans($N / 2 * $D);
    } else {
        ans(($N - 1) / 2 * $D + $lit1);
        // exit("$N $D $lit1\n");
    }
}

function ans($ans) {
    printf("%d\n", $ans);
}