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;
using ll = long long;
using Graph = vector<vector<int>>;
#define REP(i,m,n) for(int i=(int)(m);i<(int)(n);++i)
#define rep(i,n) REP(i,0,n)
#define rREP(i,m,n) for(int (i)=(n)-1; (i)>=(m);--(i))
#define all(x) (x).begin(),(x).end()
#define out(y,x,h,w) (y)<0||(x)<0||(y)>=(h)||(x)>=(w)
constexpr int INF = 1000000001;
constexpr ll mod = 1000000007;
constexpr double PI = 3.1415926535897932;

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;
}
inline void init() {
	cin.tie(nullptr); cout.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(15);
}

int main() {
	init();

	int N; cin >> N;
	vector<int>dp(N + 1, INF);
	dp[0] = 0;
	int price[] = { 6,9 };
	rep(i, N) {
		chmin(dp[i + 1], dp[i] + 1);
		rep(j, 2) {
			int x = price[j];
			int y = x;
			while (i + x <= N) {
				dp[i + x] = min(dp[i + x], dp[i] + 1);
				x *= y;
			}
		}
	}
	cout << dp[N] << "\n";

	return 0;
}