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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>

#define UPPER_BOUND 100000
#define UPPER_BOUND_INDEX 12
#define POW_BOUND 7+1

typedef enum { false, true } bool;

int change[UPPER_BOUND_INDEX];

int min = 100000;

void change_array(void)
{
    int i, index6, index9, pow6[POW_BOUND], pow9[POW_BOUND];
    
    
    index6 = index9 = 0;
    
    
    for(i=0; i<POW_BOUND; i++)
    {
        pow6[i] = pow(6, i);
        pow9[i] = pow(9, i);
    }
    
    for(i=0; pow6[index6]<=UPPER_BOUND || pow9[index9]<=UPPER_BOUND; i++)
    {
        if(pow6[index6] < pow9[index9])
        {
            change[i] = pow6[index6];
            
            index6++;
                
            continue;
        }
        if(pow6[index6] > pow9[index9])
        {
            change[i] = pow9[index9];
            
            index9++;
            
            continue;
        }
        if(pow6[index6] == pow9[index9])
        {
            change[i] = pow6[index6];
            
            index6++;
            index9++;
            
            continue;
        }
    }
    /*for(i=0; i<UPPER_BOUND_INDEX; i++)
    {
        printf("%d\n", change[i]);
    }*/
}

int change_index(int N)
{
    int i;
    
    
    for(i=UPPER_BOUND_INDEX-1; i>=0; i--)
    {
        if(change[i] <= N)
            break;;
    }
    
    return i;
}

bool DIV(int R, int index, int num)
{
    int i;
    
    
    if(num >= min)
    {
        return false;
    }
    else if(index == 0)
    {
        if(min > num + R)
            min = num + R;
            
        return true;
    }
    else
    {
        if(R >= change[index])
        {
            for(i=R/change[index]; i>=0; i--)
            {
                if(!DIV(R-change[index]*i, index-1, num+i))
                {
                    return true;
                }
            }
        }
        else
        {
            if(!DIV(R, index-1, num))
            {
                return true;
            }
        }
    }
}

int main(void)
{
    int N;
  
  
    scanf("%d", &N);


    change_array();

    DIV(N, change_index(N), 0);


    printf("%d\n", min);


    //system("pause");
    return 0;
}