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 (1.42.0) (judge verdict: Runtime Error)

use proconio::input;

fn main() {
  input!(q:usize, h:usize, s:usize, d:usize, n:usize);
  let n = n * 100;
  let h = h.min(2*q);
  let s = s.min(2*h);
  let d = d.min(2*s);
  println!("{}", solve(n, q, h, s, d));
}

fn solve(n: usize, q: usize, h: usize, s: usize, d: usize) -> usize {
  if n >= 200 {
    d + solve(n - 200, q, h, s, d)
  } else if n >= 100 {
    s + solve(n - 100, q, h, s, d)
  } else if n >= 50 {
    h + solve(n - 50, q, h, s, d)
  } else if n >= 25 {
    q + solve(n - 25, q, h, s, d)
  } else if n == 0 {
    0
  } else {
    unreachable!()
  }
}