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>
#include <sys/time.h>
using namespace std;

#define rep(i,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mt make_tuple
#define mp make_pair
template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); }
template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>;
ll ugauss(ll a, ll b) { if (!a) return 0; if (a>0^b>0) return a/b; else return (a+(a>0?-1:1))/b+1; }
ll lgauss(ll a, ll b) { if (!a) return 0; if (a>0^b>0) return (a+(a>0?-1:1))/b-1; else return a/b; }
template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) {  o << "(" << v.first << ", " << v.second << ")"; return o; }
template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq<0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, size_t... Is>
void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 0)...}; }
template<class Ch, class Tr, class... Args> 
auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args)>()); return os << ")"; }
ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; }
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T> ostream &operator<<(ostream &o, const deque<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const unordered_set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U>  ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U, typename V>  ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it; o << "]";  return o; }
vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; }
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}
template <typename T, typename S, typename U> ostream &operator<<(ostream &o, const priority_queue<T, S, U> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const queue<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.front(); tmp.pop(); o << x << " ";} return o; }
template <typename T> ostream &operator<<(ostream &o, const stack<T> &v) { auto tmp = v; while (tmp.size()) { auto x = tmp.top(); tmp.pop(); o << x << " ";} return o; }
template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;};
void vizGraph(vvll& g, int mode = 0, string filename = "out.png") { ofstream ofs("./out.dot"); ofs << "digraph graph_name {" << endl; set<P> memo; rep(i, g.size())  rep(j, g[i].size()) { if (mode && (memo.count(P(i, g[i][j])) || memo.count(P(g[i][j], i)))) continue; memo.insert(P(i, g[i][j])); ofs << "    " << i << " -> " << g[i][j] << (mode ? " [arrowhead = none]" : "")<< endl;  } ofs << "}" << endl; ofs.close(); system(((string)"dot -T png out.dot >" + filename).c_str()); }
struct timeval start; double sec() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec - start.tv_sec) + (tv.tv_usec - start.tv_usec) * 1e-6; }
size_t random_seed; struct init_{init_(){ ios::sync_with_stdio(false); cin.tie(0); gettimeofday(&start, NULL); struct timeval myTime; struct tm *time_st; gettimeofday(&myTime, NULL); time_st = localtime(&myTime.tv_sec); srand(myTime.tv_usec); random_seed = RAND_MAX / 2 + rand() / 2; }} init__;
#define ldout fixed << setprecision(40) 

#define EPS (double)1e-14
#define INF (ll)1e18
#define mo  (ll)(1e9+7)

// 素数の個数はO(n / log n)

/**********************************************************/
// 素数テーブルの作成
/**********************************************************/
// 素数テーブル構築: O(n log n)
#define MAX_NUM 20000000
ll prime_size = -1;
bool is_prime[MAX_NUM];
vector<ll> primes;      // 素数リスト
void sieve_of_atkin(ll N) {
    ll sqrtN = sqrt(N);
    int n;
    for (int z = 1; z <= 5; z += 4) {
        for (int y = z; y <= sqrtN; y += 6) {
            for (int x = 1; x <= sqrtN && (n = 4*x*x+y*y) <= N; ++x)
                is_prime[n] = !is_prime[n];
            for (int x = y+1; x <= sqrtN && (n = 3*x*x-y*y) <= N; x += 2)
                is_prime[n] = !is_prime[n];
        }
    }
    for (int z = 2; z <= 4; z += 2) {
        for (int y = z; y <= sqrtN; y += 6) {
            for (int x = 1; x <= sqrtN && (n = 3*x*x+y*y) <= N; x += 2)
                is_prime[n] = !is_prime[n];
            for (int x = y+1; x <= sqrtN && (n = 3*x*x-y*y) <= N; x += 2)
                is_prime[n] = !is_prime[n];
        }
    }
    for (int y = 3; y <= sqrtN; y += 6) {
        for (int z = 1; z <= 2; ++z) {
            for (int x = z; x <= sqrtN && (n = 4*x*x+y*y) <= N; x += 3)
                is_prime[n] = !is_prime[n];
        }
    }
    for (int n = 5; n <= sqrtN; ++n)
        if (is_prime[n])
            for (int k = n*n; k <= N; k+=n*n)
                is_prime[k] = false;
    is_prime[2] = is_prime[3] = true;
}

void sieve_of_eratosthenes(ll n) {
    for (ll i = 2; i < n; ++i)
        is_prime[i] = true;
    for (ll i = 2; i*i < n; ++i)
        if (is_prime[i])
            for (ll j = i*i; j < n; j+=i)
                is_prime[j] = false;
    for (ll i = 0; i < n; i++) 
        if (is_prime[i]) 
            primes.push_back(i);
}

// n以下の素数リストを作成する
//
// 素数はn個でも、n^2までの素因数分解ができる！
// よほどのことがなければn=1e6+10とかを確保すれば十分。
//
// 1e7は割とMLEするので注意
void constructPrime(ll n) {
    if (n > 5e6) {
        cerr << "WATCH OUT!!!! YOU ARE USING TOO LARGE MEMORY!" << endl;
    }
    assert(n < MAX_NUM);
    prime_size = n;
#if 1
    sieve_of_atkin(n);
#else
    sieve_of_eratosthenes(n);
#endif
    primes.reserve(1.2*n*log(n)); // 素数定理よりn>100でOK
    rep(i, n) if (is_prime[i]) {
        primes.pb(i);
    }
}

/**********************************************************/
// 以下はよく行う処理のテンプレート
// そのままは大抵使えないので、改造するつもりで。
/**********************************************************/
// 素因数分解
//
// 小さい素数から見ていって、割れたら割って素因数リストに追加する
// 残ったnが素数リストよりも大きければ、そのnは素数だと見なして返す。
unordered_map<ll, ll> factorize(ll n) {
    if (n <= 1) return unordered_map<ll, ll>();
    unordered_map<ll, ll> divisors_list;

    ll prime;
    for (ll i = 0; (prime = primes[i]) && n >= prime * prime; )  
        if (n % prime) 
            i++;
        else 
            divisors_list[prime]++, n /= prime;
    if (n != 1) 
        divisors_list[n]++;
    return divisors_list;
}

// [0, n]の範囲を全て素因数分解する。「飛び飛びの」高速化。結構速い。
// n < 1,000,000で150ms
// n < 10,000,000で1000ms 
//
// 配列rem[i]は、初めiが入っている。
// 素数pを下から見るが、rem[i]を全て見る必要はなく、rem[p*i]だけでいい。
vector<unordered_map<ll, ll>> factorizeRange(ll n) {
    vector<unordered_map<ll, ll>> fact(n+1);

    vector<ll> rem(n+1); rep(i, n+1) rem[i] = i;
    for (auto x : primes) for (ll i = x; i <= n; i += x) 
        while (rem[i] % x == 0) 
            rem[i] /= x, fact[i][x]++;
    return fact;
}


// lcm(a)
// a[i] < 1e7
unordered_map<ll, ll> lcmSmall(set<ll>& a) {
    if (!a.size()) return unordered_map<ll, ll>();

    ll n = *(prev(a.end()));
    vector<ll> rem(n+1); 
    rep(i, n+1) rem[i] = i;
    unordered_map<ll, ll> ret;
    for (auto x : primes) {
        ll max_c = 0;
        for (ll i = x; i <= n; i += x) if (a.count(i)) {
            ll c = 0;
            while (rem[i] % x == 0) 
                rem[i] /= x, c++;
            chmax(max_c, c);
        }
        if (max_c) 
            ret[x] = max_c;
    }
    return ret;
}

// lcm(a)
// a[i]>1e7
unordered_map<ll, ll> lcmLarge(set<ll>& a) {
    if (!a.size()) return unordered_map<ll, ll>();

    unordered_map<ll, ll> ret;
    for (auto n : a) {
        auto d = factorize(n);
        for (auto x : d) 
            chmax(ret[x.fi], x.se);
    }
    return ret;
}


// 約数を全列挙する。
//
// 約数を試し割りするより、n=1000000で4倍くらい早い。
vector<ll> divisors(ll n) {
    vector<ll> divisors_list;

    auto counter = factorize(n);
    divisors_list.push_back(1);
    for (auto x : counter) {
        ll tmp_size = divisors_list.size();
        ll p = 1;
        for (ll i = 0; i < x.second; i++) {
            p *= x.first;
            for (ll j = 0; j < tmp_size; j++) 
                divisors_list.push_back(divisors_list[j] * p);
        }
    }
    return divisors_list;
}

ll getDivisorsNum(ll n) {
    unordered_map<ll, ll> divisors_list = factorize(n);
    ll p = 1;
    for (auto x : divisors_list) 
        p *= x.second + 1;
    return p;
}


/**********************************************************/
// 前処理なしの素数判定
/**********************************************************/
using u64 = uint32_t;
using u128 = uint64_t;
u128 mul(u64 a, u64 b, u64 m) { return u128(a) * b % m; };
bool isPrime(unsigned long long n) {
    if (n <= 1) return false;
    if (n == 2) return true;
    static mt19937 mt(time(NULL));
    u64 s = n - 1;
    int e = 0;
    for (; s % 2 == 0; s /= 2) e++;
    for (int ii = 0; ii < 8; ii++) {
        u64 x = std::uniform_int_distribution<u64>(2, n - 1)(mt);
        u64 r = 1;
        for (u64 i = s; i > 0; i >>= 1, x = mul(x, x, n)) {
            if (i & 1) r = mul(r, x, n);
        }
        if (r == 1) continue;
        for (int i = 1; i < e && r != n - 1; i++) {
            r = mul(r, r, n);
        }
        if (r != n - 1) return false;
    }
    return true;
}


// O(n^0.25)
// ローのアルゴリズム
// 計算結果はrho_retに保存される
using ull = unsigned long long;
map<ull,int> rho_ret; // {p, k}, pの素因数がk個存在
ull find_factor(ull z) {
    if (!(z&1)) return 2;
    ull c = rand() % z, x = 2, y = 2, d = 1;
    while (d == 1) {
        ull tp = (mul(y,y,z) + c) % z;
        y = (mul(tp,tp,z) + c) % z;
        x = (mul(x,x,z) + c) % z;
        d = __gcd(x>y ? x-y : y-x, z);
    }
    return d;
}
void rhofact_rec(ull z) {
    if (z==1) return;
    if (isPrime(z)) { rho_ret[z]++; return; }
    while(1) {
        ull f = find_factor(z);
        if (f != z) {
            rhofact_rec(f);
            rhofact_rec(z/f);
            break;
        }
    }
}
void rho(ull z) {
    rho_ret.clear();
    rhofact_rec(z);
}

// ガウス素数＝複素数の素数判定
bool isGaussianPrime(ll a, ll b) {
    if (a < 0) a = -a;
    if (b < 0) b = -b;
    if (a == 0) return b % 4 == 3 && is_prime[b];
    if (b == 0) return a % 4 == 3 && is_prime[a];
    return is_prime[a*a+b*b];
}

// 区間篩
// O( n log n )．
const ll N = 100000000; // MAXPRIME 
const ll M = 10000;     // SQRT(N)
const ll K = 6000000;   // NUMBER OF PRIMES, CHOOSE 9/8 * N / LOG(N)
vector<ll> iterativeSieve() {
    static ll p[K], table[M];
    for (ll i = 2; i < M; ++i) p[i] = i;
    for (ll i = 2; i*i < M; ++i)
        if (p[i])
            for (ll j = i*i; j < M; j += i)
                p[j] = 0;
    p[0] = p[1] = 0;
    ll num = remove(p, p+M, 0) - p;
    for (ll m = M; m < N; m += M) {
        for (ll x = m; x < m+M; ++x)
            table[x-m] = x;
        for (ll i = 0, j; p[i]*p[i] < m+M; ++i) {
            if (p[i] >= m)          j = p[i] * p[i];
            else if (m % p[i] == 0) j = m;
            else                    j = m - (m % p[i]) + p[i];
            for (; j < m+M; j += p[i]) table[j-m] = 0;
        }
        num = remove_copy(table, table+M, p+num, 0) - p;
    }
    return vector<ll>(p, p+num);
}


int main(void) {
    ll n; cin >> n;
    vll cand = {1};
    ll tmp;

    tmp = 6;
    while (tmp < 200000) {
        cand.pb(tmp);
        tmp *= 6;
    }
    tmp = 9;
    while (tmp < 200000) {
        cand.pb(tmp);
        tmp *= 9;
    }

    vector<ll> dp(n+1, INF);
    dp[0] = 0;
    rep(i, n+1) {
        for (auto x : cand) {
            if (i + x <= n) {
                chmin(dp[i+x], dp[i]+1);
            }
        }
    }
    cout << dp[n] << endl;


    return 0;
}