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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include <algorithm>
#include <functional>
using namespace std;
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(var,init,end) for(int var=init;var<end;var++)
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
int facctorialMethod(int k) {
	int sum = 1;
	for (int i = 1; i <= k; ++i)
	{
		sum *= i;
	}
	return sum;
}

int pow(int a, int b) {
	int tmp = a;
	FOR(i, 0, b-1) {
		a *= tmp;
	}
	return a;
}

/*
int main() {
	int n, c;
	cin >> n >> c;

	vector<vector <int>>d;
	d = vector<vector<int>>(c, vector<int>(c, 0));
	FOR(i, 0, c) {
		FOR(j, 0, c) {
			cin >> d[i][j];
		}
	}

	vector<vector <int>>mat;
	mat = vector<vector<int>>(n, vector<int>(n, 0));
	FOR(i, 0, n) {
		FOR(j, 0, n) {
			cin >> mat[i][j];
		}
	}
	int cnt = 0;

	cout << cnt;
	return 0;
}
*/


int main() {
	int n;
	int lower[35] = {
		1,2,3,4,5,
		1,2,3,1,2,3,
		2,3,4,2,3,4,2,3,4,
		3,4,5,3,4,5,3,4,5,
		4,5,6,4,5,6 };
	cin >> n;

	vector<int> nums(12);
	nums[0] = 1;
	FOR(i, 0, 6) {
		nums[i + 1] = pow(6, (i + 1));
	}
	FOR(i, 0, 5) {
		nums[i + 7] = pow(9, (i + 1));
	}

	sort(nums.begin(), nums.end(), greater<int>());
	int cnt = 0;
	while (n > 0) {
		if (n <= 35) {
			cnt += lower[n - 1];
			break;
		}
		for (int i = 0; i < 12; i++) {
			if (n >= nums[i]) {
				n -= nums[i];
				cnt++;
				break;
			}
		}
	}
	cout << cnt;
	return 0;
}