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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author mikit
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CStrangeBank solver = new CStrangeBank();
        solver.solve(1, in, out);
        out.close();
    }

        static int[] DP = new int[100000 + 1];
    static class CStrangeBank {
        static final int[] C6 = {6, 36, 216, 1296, 7776, 46656};
        static final int[] C9 = {9, 81, 729, 6561, 59049};
        int[] m;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            m = new int[n + 1];
            out.println(min(n));
        }

        int min(int n) {
            if (n == 0) {
                return 0;
            }
          if(DP[n]!=0){return DP[n];}
            int min = Integer.MAX_VALUE;
            for (int i = C6.length - 1; i >= 0; i--) {
                if (n >= C6[i]) {
                    min = Math.min(min, min(n - C6[i]) + 1);
                    //break;
                }
            }
            for (int i = C9.length - 1; i >= 0; i--) {
                if (n >= C9[i]) {
                    min = Math.min(min, min(n - C9[i]) + 1);
                    //break;
                }
            }
            min = Math.min(min,  min(n - 1) + 1);
          return DP[n]=min;
        }

    }
}