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>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rrep(i,n) for (int i=(int)(n)-1; i>=0; --i)
#define debug(x) cerr << #x << " = " << x << "\n"
#define debugv(x) rep(i, x.size()) cerr << x[i] << (i == x.size() - 1 ? "\n" : " ")
#define all(x) x.begin(),x.end()
#define int long long
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = ((1LL << 30) + (1LL << 29));
const int MOD = 1000000007;

//=============================
int N,M,K;
string S,T;
vector<int> A;
vector<int> B;
//=============================
void input()
{
    cin>>N;
}
//=============================
void func(){}
//=============================
void solve()
{
    A.resize(30);
    B.resize(30);
    A[0]=1;
    B[0]=1;
    rep(i,29){
        A[i+1]=A[i]*6;
        B[i+1]=B[i]*9;
    }

    int ans;
    for (ans=0; N ; ++ans) {
        auto a = lower_bound(all(A),N);
        auto b = lower_bound(all(B),N);
        if (*a == N || *b == N) {
            --N;
        }
        else {
            --a;--b;
            N -= max(*a,*b);
        }
    }

    cout<<ans<<endl;


    
}
//=============================
signed main()
{
    input();
    solve();
    return 0;
}