#include <cstdio>
#include <iostream> 
using namespace std;

int n;
int main()
{
	while(~scanf("%d" ,&n))
	{
		int res=n; //最大情况肯定为n次  全部每次取1元 
		for(int i=0;i<=n;i++)
		{
			int cc=0;
			int t=i;
			while(t) cc+=t%6,t/=6;  //计算6进制的系数和 
			t=n-i;
			while(t) cc+=t%9,t/=9;  //计算9进制的系数和 
			if(res>cc) res=cc; 
		
		}
		printf("%d\n",res);
	}

}

### Prompt

Create a solution in cpp for the following problem:

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.

Examples

Input

127


Output

4


Input

3


Output

3


Input

44852


Output

16

### Response

```cpp
#include <cstdio>
#include <iostream> 
using namespace std;

int n;
int main()
{
	while(~scanf("%d" ,&n))
	{
		int res=n; //最大情况肯定为n次  全部每次取1元 
		for(int i=0;i<=n;i++)
		{
			int cc=0;
			int t=i;
			while(t) cc+=t%6,t/=6;  //计算6进制的系数和 
			t=n-i;
			while(t) cc+=t%9,t/=9;  //计算9进制的系数和 
			if(res>cc) res=cc; 
		
		}
		printf("%d\n",res);
	}

}

```