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 "bits/stdc++.h"
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(x) (x).begin(), (x).end()  //昇順
#define RALL(x) (x).rbegin(), (x).rend() // 降順
#define SZ(x) ((int)(x).size)
const long long mod = 1e9+7;
typedef priority_queue<int> PQ;
typedef vector<long long> Vl;
typedef vector<int> VI;  // VI a(n);
typedef vector<string> VS;  
typedef vector<VS> VSS;  
typedef vector<VI> VVI;   // VVI a(n,vector<int>(m)) n * m 
typedef vector<Vl> VVl;
typedef pair<int, int> PII;
typedef vector<pair<int, int> > PS;
typedef long long ll;  //ll とdoubleは違う
// cout << << endl;
// cin >> ;
template<typename T>
T gcd(T a, T b){
	return b != 0 ? gcd(b, a % b) : a;
}    
template<typename T>
T lcm(T a, T b){
    return a/gcd(a,b)*b;
}       

int main(){
    ll c=0,d=0,di=0,n,h,w,l,r,min=0,ptr=0;
    string s,s1,s2;
    cin >> n;
    min = n;
    rep(i,n){
        c=0;
        di=i;
        while(di>0)
            c += di%6 , di /= 6;
        di = n - i;
        while(di>0)
            c += di%9 , di /= 9;

        if(min > c)
            min = c;
    }
    cout << min << endl;
    return 0;
}