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: Accepted)

#include "bits/stdc++.h"
using namespace std;

#define ASC(vec) vec.begin(), vec.end()		// 昇順ソート 例：sort(ASC(vec));
#define DESC(vec) vec.rbegin(), vec.rend()	// 降順ソート 例:sort(DESC(vec));
#define rep(i, n) for(i = 0; i < n; i++)
#define Rep(i, n) for(i = 1; i < n; i++)

const int inf = (int)1e9 + 7;
const long long INF = 1LL << 60;

template<class T> inline bool chmin(T & a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

typedef long long ll;

// 4,8方向に探索する際に用いる配列
//int dx[4] = { 0,  0, 1, -1 };
//int dy[4] = { 1, -1, 0,  0 };
//int dx[8] = { 1, 0, -1,  0, 1, -1, -1,  1 };
//int dy[8] = { 0, 1,  0, -1, 1,  1, -1, -1 };

int dp[100010];

int main()
{
	int n;
	cin >> n;

	vector<int> nums(1, 1);

	for (int i = 6; i <= n; i *= 6)
	{
		nums.push_back(i);
	}
	for (int i = 9; i <= n; i *= 9)
	{
		nums.push_back(i);
	}

	sort(ASC(nums));

	for (int i = 0; i < 100010; i++) dp[i] = inf;

	dp[0] = 0;

	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j < nums.size(); j++)
		{
			if (i - nums[j] < 0) continue;

			chmin(dp[i], dp[i - nums[j]] + 1);
		}
	}

	//for (int i = 0; i <= n; i++) cout << dp[i] << " ";

	cout << dp[n] << endl;
}