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 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(var, begin, end) for (int var = (begin); var < (end); var++)
#define RFOR(var, begin, end) for (int var = (begin) - 1; var >= (end); var--)
#define REP(var, length) FOR(var, 0, length)
#define RREP(var, length) RFOR(var, length, 0)
#define SORT(vector) sort(vector.begin(), vector.end())

const ll MOD = 1e9 + 7;
const int MAX = 1e6 + 1;

vector<int> memo;
vector<int> amt;

int func(int n) {
  if (memo[n] > 0) {
    return memo[n];
  }
  if (n == 0) {
    return 0;
  }
  int ret = 1e9;
  REP(i, amt.size()) {
    if (amt[i] > n) {
      continue;
    }
    ret = min(ret, func(n - amt[i]) + 1);
  }
  memo[n] = ret;
  return ret;
}

void solve(istream& cin, ostream& cout) {
  int n;
  cin >> n;
  memo = vector<int>(n + 1, -1);
  set<int> amount;

  for (int i = 1; i < MAX; i *= 6) {
    amount.insert(i);
  }

  for (int i = 1; i < MAX; i *= 9) {
    amount.insert(i);
  }

  for (set<int>::iterator itr = amount.begin(); itr != amount.end(); itr++) {
    amt.push_back(*itr);
  }

  cout << func(n) << endl;
}

#ifndef TEST
int main() {
  solve(cin, cout);
}
#endif