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: Accepted)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let it = stdin.lock().bytes().map(|c| c.unwrap() as char);
    let (fs, ft) = (|c: &char| c.is_whitespace(), |c: &char| !c.is_whitespace());
    let token = it.skip_while(fs).take_while(ft).collect::<String>();
    token.parse().ok().unwrap()
}
#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    f(std::io::BufWriter::new(out.lock()));
}
#[allow(unused_macros)]
macro_rules! debug {
    ($($var:expr),*) => { #[cfg(debug_assertions)] {
        $({ match &$var { v => eprint!("{}={:?} ", stringify!($var), v) } })*
        eprintln!();
    }};
}

fn main() {
    let n: i32 = input();

    let p6: Vec<i32> = (1..)
        .map(|i| i32::pow(6, i))
        .take_while(|&i| i <= n)
        .collect();
    let p9: Vec<i32> = (1..)
        .map(|i| i32::pow(9, i))
        .take_while(|&i| i <= n)
        .collect();

    let mut dp = vec![i32::max_value(); n as usize + 1];
    dp[0] = 0;
    for i in 1..n as usize + 1 {
        dp[i] = std::cmp::min(dp[i], dp[i - 1] + 1);
        for &j in &p6 {
            let j = j as usize;
            if i < j {
                break;
            }
            dp[i] = std::cmp::min(dp[i], dp[i - j] + 1);
        }
        for &j in &p9 {
            let j = j as usize;
            if i < j {
                break;
            }
            dp[i] = std::cmp::min(dp[i], dp[i - j] + 1);
        }
    }
    println!("{}", dp[n as usize]);
}