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 Python (3.4.3) (judge verdict: Accepted)

import math
import fractions
import bisect
import collections
import itertools
import heapq
import string
import sys
import copy
from decimal import *
from collections import deque
sys.setrecursionlimit(10**7)
MOD = 10**9+7
INF = float('inf') #無限大
def gcd(a,b):return fractions.gcd(a,b) #最大公約数
def lcm(a,b):return (a*b) // fractions.gcd(a,b) #最小公倍数
def iin(): return int(sys.stdin.readline()) #整数読み込み
def ifn(): return float(sys.stdin.readline()) #浮動小数点読み込み
def isn(): return sys.stdin.readline().split() #文字列読み込み
def imn(): return map(int, sys.stdin.readline().split()) #整数map取得
def imnn(): return map(lambda x:int(x)-1, sys.stdin.readline().split()) #整数-1map取得
def fmn(): return map(float, sys.stdin.readline().split()) #浮動小数点map取得
def iln(): return list(map(int, sys.stdin.readline().split())) #整数リスト取得
def iln_s(): return sorted(iln()) # 昇順の整数リスト取得
def iln_r(): return sorted(iln(), reverse=True) # 降順の整数リスト取得
def fln(): return list(map(float, sys.stdin.readline().split())) # 浮動小数点リスト取得
def join(l, s=''): return s.join(l) #リストを文字列に変換
def perm(l, n): return itertools.permutations(l, n) # 順列取得
def perm_count(n, r): return math.factorial(n) // math.factorial(n-r) # 順列の総数
def comb(l, n): return itertools.combinations(l, n) # 組み合わせ取得
def comb_count(n, r): return math.factorial(n) // (math.factorial(n-r) * math.factorial(r)) #組み合わせの総数
def two_distance(a, b, c, d): return ((c-a)**2 + (d-b)**2)**.5 # 2点間の距離
def m_add(a,b): return (a+b) % MOD
def print_list(l): print(*l, sep='\n')
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def No(): print('No')

def sieves_of_e(n):
    is_prime = [True] * (n+1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, int(n**0.5)+1):
        if not is_prime[i]: continue
        for j in range(i * 2, n+1, i): is_prime[j] = False
    return is_prime

N = iin()


money = []
pow6 = 1
while pow6 <= N:
    money.append(pow6)
    pow6 *= 6
pow9 = 9
while pow9 <= N:
    money.append(pow9)
    pow9 *= 9
lm = len(money)
dp = [[INF for _ in range(N+1)] for _ in range(lm+1)]
dp[0][0] = 0
for i in range(lm):
    for j in range(N+1):
        dp[i+1][j] = dp[i][j]
        if j - money[i] >= 0:
            dp[i+1][j] = min(dp[i+1][j], dp[i+1][j-money[i]]+1)
print(dp[lm][N])