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 Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let (Q, H, S, D): (i64, i64, i64, i64) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };
    let N: i64 = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };

    let c = vec![8 * Q, 4 * H, 2 * S, D];
    let m = (0..4).min_by_key(|&i| c[i]).unwrap();

    let ans = if m == 0 {
        4 * Q * N
    } else if m == 1 {
        2 * H * N
    } else if m == 2 {
        S * N
    } else {
        let x = *vec![4 * Q, 2 * H, S].iter().min().unwrap();
        (N / 2) * D + x
    };

    println!("{}", ans);
}