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;
#define ll          long long int
#define pb          push_back
#define pll         pair<ll,ll>
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mod         1000000007
#define inf         1000000000000000000LL
#define vl          vector<ll>
#define F           first
#define S           second
#define db          long double
#define sz(x)       (ll)x.size()
#define fix(n)      cout<<fixed<<setprecision(n)
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define all(x)      x.begin(),x.end()
#define pi          3.14159265358979323

ll n;
ll ps[10], pn[10];
ll moves;


void pre()
{
       ps[0]=1;
       pn[0]=1;
       for(ll i=1; i<10; i++) ps[i] = ps[i-1]*6;
       for(ll i=1; i<10; i++) pn[i] = pn[i-1]*9;
}


void brute(ll num, ll steps)
{
       if(num==0) { moves=min(moves, steps); return; }
       for(ll i=0; i<10; i++)
       {
              if(ps[i] > num)
              {
                     brute(num - ps[i-1], steps+1);
                     break;
              }
       }
       for(ll i=0; i<10; i++)
       {
              if(pn[i] > num)
              {
                     brute(num - pn[i-1], steps+1);
                     break;
              }
       }
}

void solve()
{
       cin>>n;
       moves=n;
       pre();
       brute(n, 0);
       cout<<moves<<endl;
}

int main()
{
       speed_up;
       ll T=1;
       //cin>>T;
       while(T--) solve();
	return 0;
}