Problem:
Score : 300 points

Problem Statement

To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:

1 yen (the currency of Japan)

6 yen, 6^2(=36) yen, 6^3(=216) yen, ...

9 yen, 9^2(=81) yen, 9^3(=729) yen, ...

At least how many operations are required to withdraw exactly N yen in total?

It is not allowed to re-deposit the money you withdrew.

Constraints

1 \leq N \leq 100000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If at least x operations are required to withdraw exactly N yen in total, print x.

Sample Input 1

127

Sample Output 1

4

By withdrawing 1 yen, 9 yen, 36(=6^2) yen and 81(=9^2) yen, we can withdraw 127 yen in four operations.

Sample Input 2

3

Sample Output 2

3

By withdrawing 1 yen three times, we can withdraw 3 yen in three operations.

Sample Input 3

44852

Sample Output 3

16

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

// Rustで競技プログラミングの入力をスッキリ記述するマクロ
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

use std::cmp::Ordering;

fn solve(n: u32) {
    fn calc(count: u32, left: u32, exp6: u32, exp9: u32) -> u32 {
        let pow6 = u32::pow(6, exp6);
        let pow9 = u32::pow(9, exp9);
        match (left, left.cmp(&pow6), left.cmp(&pow9), pow6.cmp(&pow9)) {
            (left, _, _, _) if left <= 5 => count + left,
            (left, Ordering::Greater, Ordering::Less, _)
            | (left, Ordering::Greater, Ordering::Greater, Ordering::Greater) => {
                calc(count + 1, left - pow6, exp6, exp9)
            }
            (left, Ordering::Less, Ordering::Greater, _)
            | (left, Ordering::Greater, Ordering::Greater, Ordering::Less) => {
                calc(count + 1, left - pow9, exp6, exp9)
            }
            (left, _, _, _) => calc(count, left, exp6 - 1, exp9 - 1),
        }
    }
    let ans = calc(0, n, 6, 5);
    println!("{}", ans);
}

fn main() {
    input! {
        n: u32
    };
    solve(n);
}