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)

use std::cmp::*;

#[allow(unused_macros)]
macro_rules! get {
    ([$t:ty] ; $n:expr) => {
        (0..$n)
        .map(|_| get!([$t])).collect::<Vec<_>>()
    };
    ($($t:ty),+ ; $n:expr) => {
        (0..$n).map(|_| get!($($t),+)).collect::<Vec<_>>()
    };
    ([$t:ty]) => {
        rl().split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect::<Vec<$t>>()
    };
    ($t:ty) => {
        rl().parse::<$t>().unwrap()
    };
    ($($t:ty),*) => {
        {
            let line =rl();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap()),*)
        }
    };
}
fn rl() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn main() {
    let n = get!(usize);
    let mut dp = vec![1 << 20; n + 1];
    dp[1] = 1;
    for i in 1..n + 1 {
        let mut j = 1;
        while i + j < n + 1 {
            dp[i + j] = min(dp[i + j], dp[i] + 1);
            j *= 6;
        }
        j = 1;
        while i + j < n + 1 {
            dp[i + j] = min(dp[i + j], dp[i] + 1);
            j *= 9;
        }
    }
    println!("{}", dp[n]);
}