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)

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            line.pop();
            line
        }
    )
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let iter = line.split_whitespace();
            iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    )
}

fn solve(n: usize) -> usize {
    let mut dp = vec![std::usize::MAX; n+1];

    dp[0] = 0;

    for i in 0..n+1 {
        let v = dp[i];
        for j in 0.. {
            let total = i + 6usize.pow(j);
            if total > n {
                break
            }
            if v+1 < dp[total] {
                dp[total] = v+1;
            }
        }

        for j in 0.. {
            let total = i + 9usize.pow(j);
            if total > n {
                break
            }
            if v+1 < dp[total] {
                dp[total] = v+1;
            }
        }
    }

    dp[n]
}

fn main() {
    let n = parse_line!(usize);

    println!("{}", solve(n));
}

#[test]
fn solve_test() {
    assert_eq!(solve(127), 4);
    assert_eq!(solve(3), 3);
    assert_eq!(solve(44852), 16);
    assert_eq!(solve(1), 1);
}