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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Reflection.Metadata;
using System.Runtime.CompilerServices;
using System.Runtime.InteropServices;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading;

namespace AtCoder
{
	public class Program
	{
		static void Main()
		{
			long n = long.Parse(Console.ReadLine());

			var monay = new List<long>();
			monay.Add(1);

			long temp = 6;
			while (temp <= n) {
				monay.Add(temp);
				temp *= 6;
			}

			temp = 9;
			while (temp <= n) {
				monay.Add(temp);
				temp *= 9;
			}

			var dp = new long[n + 1];
			for (int i = 0; i < n + 1; i++) {
				dp[i] = long.MaxValue;
			}

			dp[0] = 1;
			foreach (var m in monay) {
				dp[m] = 1;
			}

			for (int i = 0; i < n; i++) {
				if (dp[i] == long.MaxValue) {
					continue;
				}

				foreach (var m in monay) {
					long next = i + m;
					if (next <= n) {
						dp[next] = Math.Min(dp[next], dp[i] + 1);
					}
				}
			}

			long ans = dp[n];
			Console.WriteLine(ans);
		}
	}
}