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 REP(i,n) FOR(i,0,n)
#define FOR(i,a,b) for(int i=(a),i##Len_=(b);i<i##Len_;i++)
#define ALL(x) (x).begin(),(x).end()
template <class T> void chmin(T& a, const T& b) { if (b < a) a = b; }
const char* LF = "\n";
namespace{struct _{_(){
    cin.tie(0);
    ios::sync_with_stdio(false);
}}a;}

vector<int> C;
vector<int> memo;

int find(int n) {
    if (n <= 0) return 0;
    if (memo[n] != -1) return memo[n];

    int ret = n;
    REP(i, C.size()) {
        if (n - C[i] >= 0) chmin(ret, find(n - C[i]) + 1);
    }

    return memo[n] = ret;
}

signed main() {
    int N;
    cin >> N;
    memo.resize(N+1, -1);

    C.reserve(100);

    FOR(i, 1, 10) {
        if (pow(6, i) <= N) C.push_back(pow(6, i));
        if (pow(9, i) <= N) C.push_back(pow(9, i));
    }
    C.push_back(1);

    cout << find(N) << LF;
    return 0;
}