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::{stdin, Read};
use std::str::FromStr;

// https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

#[allow(dead_code)]
fn main() {
    let n = read::<usize>();
    let inf = usize::max_value() as usize;
    let mut dp = vec![inf; n + 1];
    dp[0] = 0;

    use std::cmp::min;

    for i in 1..n + 1 {
        // 1
        if dp[i - 1] != inf {
            dp[i] = min(dp[i], dp[i - 1] + 1)
        }
        // 6
        let mut six = 6;
        while six <= i {
            let index = i - six;
            if dp[index] != inf {
                dp[i] = min(dp[i], dp[index] + 1)
            }
            six *= 6;
        }

        // 9
        let mut nine = 9;
        while nine <= i {
            let index = i - nine;
            if dp[index] != inf {
                dp[i] = min(dp[i], dp[index] + 1)
            }
            nine *= 9;
        }
    }

    println!("{}", dp[n])
}