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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define REP(i,n) for (int64_t i=0; i<(n); ++i)
#define P pair<int,int>
using ll=int64_t;
using namespace std;
#define ketasuu(n) fixed<<setprecision(n)
#define btoe(p) p.begin(),p.end()
#define etob(p) p.rbegin(),p.rend()


const vector<int> che={6,9};

int main(){
    int n; cin>>n;
    int ans=0;
    vector<int> p(n+5,-1);
    p[0]=0;
    rep(i,n){
        vector<int> cnt;
        rep(j,2){
            int a=che[j];
            rep(k,8){
                if((i+1)<pow(a,(k+1))) break;
                if(p[i+1-pow(a,k+1)]!=-1) cnt.push_back(p[i+1-pow(a,k+1)]+1);
            }
        }
        if(p[i]!=-1) cnt.push_back(p[i]+1);
        if(!cnt.empty()){
          sort(btoe(cnt));
          p[i+1]=cnt[0];
        }
    }
    cout<<p[n]<<endl;
}