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)

// AtCoder My Practice
// author: Leonardone @ NEETSDKASU

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!(usize);
    let mut xs = vec![1];
    let mut nine = 1;
    for i in 0..100 {
    	nine *= 9;
        if nine > n {
        	break;
        }
        xs.push(nine);
    }
    let mut six = 1;
    for i in 0..100 {
    	six *= 6;
        if six > n {
        	break;
        }
        xs.push(six);
    }
    xs.sort();
    let mut dp = vec![n+1; n+1];
    dp[0] = 0;
    loop {
    	for i in 0..dp.len() {
        	let i = dp.len()-i-1;
            if dp[i] > n {
            	continue;
            }
            for &x in xs.iter() {
            	let p = i+x;
                if p > n {
                	break;
                }
                if dp[i]+1<dp[p] {
                	dp[p] = dp[i]+1;
                }
            }
        }
        if dp[n] <= n {
        	break;
        }
    }
    let ans = dp[n];
	println!("{}", ans);    
}