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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
typedef vector<double> VD;

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

int main() {
    int n;
    cin >> n;
    VI d(n+1, -1);
    queue<int> que;

    VI next;
    next.push_back(1);
    int x = 6;
    while (x <= n){
        next.push_back(x);
        x *= 6;
    }
    x = 9;
    while (x <= n){
        next.push_back(x);
        x *= 9;
    }

    d[0] = 0;
    que.push(0);
    while (!que.empty()){
        int x = que.front();
        que.pop();
        for (int y : next){
            if (x + y > n) continue;
            if (d[x + y] != -1) continue;
            d[x + y] = d[x] + 1;
            que.push(x + y);
        }
    }
    cout << d[n] << endl;

    return 0;
}