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<stdio.h>
    //#include <bits/stdc++.h>
    #include<vector>
    #include<float.h>
    #include<iomanip>
    #include<algorithm>
    #include<string>
    #include<cstring>
    #include<math.h>
    #include<sstream>
    #include<set>
    #include<map>
    #include<queue>
    #include <cassert>
    #include <cmath>

    #define INF 1e9
    #define rep(i,n)for(int i=0;(i)<(int)(n);i++)
    #define REP(i,a,b)for(int i=(int)(a);(i)<=(int)(b);i++)
    #define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
    #define vec(type,n) vector<type>(n)
    #define vvec(m,n) vector<vector<int>> (int(m),vector<int>(n))

    using namespace std;
    using ll = long long;
    using Graph = vector<vector<int>>;


    vector<bool> seen;
    void dfs(const Graph &G,int v){
        int cnt = 0;
        seen[v] = true;
        for(auto nv:G[v]){
            if(seen[nv])continue;
            cnt++;
            dfs(G,nv);
        }
    }
    int main(){
      int n;cin>>n;
      int dp[n+1000];
      rep(i,n+1000)dp[i] = INF;
      dp[0] = 0;
      REP(i,1,n){
          for(int j=1;j<=i;j*=6){
              if(i>=j)dp[i] = min(dp[i],dp[i-j]+1);
          }
          for(int k=1;k<=i;k*=9){
              if(i>=k)dp[i] = min(dp[i],dp[i-k]+1);
          }
          //cout<<dp[i]<<endl;
      }
      cout<<dp[n]<<endl;
    }