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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

//math
//-------------------------------------------
template<class T> inline T sqr(T x) { return x * x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

VI c = { 1,9,
			9 * 9,
			9 * 9 * 9,
			9 * 9 * 9 * 9,
			9 * 9 * 9 * 9 * 9,
			9 * 9 * 9 * 9 * 9 * 9,
	6,
	6 * 6,
	6 * 6 * 6,
	6 * 6 * 6 * 6,
	6 * 6 * 6 * 6 * 6,
	6 * 6 * 6 * 6 * 6 * 6,
	6 * 6 * 6 * 6 * 6 * 6 * 6
};
const int inf = 1000000;

int main()
{
	int n;
	cin >> n;
	VVI dp = VVI(c.size()+1);
	REP(i, c.size()+1) {
		dp[i] = VI(n+1);
		REP(j, n+1)
		{
			dp[i][j] = inf;
		}
	}

	dp[0][0] = 0;
	REP(i, c.size())
	{
		REP(j, n+1)
		{
			if (j < c[i]) dp[i + 1][j] = dp[i][j];
			else {
				dp[i + 1][j] = min(dp[i][j], dp[i+1][j - c[i]] + 1);
			}
		}
	}
	cout << dp[c.size()-1][n] << endl;
	return 0;
}