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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.util.concurrent.LinkedBlockingDeque
import kotlin.math.min

fun main() {
    val n = readLine()?.toInt() ?: return
    val r = n
    var c = Int.MAX_VALUE

    fun sub(v: Int): Pair<Int, Int> {
        var six = 6
        if (six > v) {
            six = -1
        } else {
            while (true) {
                if (six == v) {
                    break
                }
                val newSix = six * 6
                if (newSix > v) {
                    break
                }
                six = newSix
            }
        }
        var nine = 9
        if (nine > v) {
            nine = -1
        } else {
            while (true) {
                if (nine == v) {
                    break
                }
                val newNine = nine * 9
                if (newNine > v) {
                    break
                }
                nine = newNine
            }
        }
        return Pair(six, nine)
    }

    val f = sub(r)
    val queue = LinkedBlockingDeque<List<Int>>()
    if (f.first != -1) {
        queue.offer(listOf(f.first))
    } else {
        c = min(c, r)
    }
    if (f.second != -1) {
        queue.offer(listOf(f.second))
    } else {
        c = min(c, r)
    }

    while (queue.isNotEmpty()) {
        val q = queue.poll()
        val s = q.sum()
        val next = sub(r - s)
        if (next.first == -1) {
            c = min(
                c, q.size + if (s == r) 0 else {
                    r - s
                }
            )
        } else {
            queue.offer(q + listOf(next.first))
        }
        if (next.second == -1) {
            c = min(
                c, q.size + if (s == r) 0 else {
                    r - s
                }
            )
        } else {
            queue.offer(q + listOf(next.second))
        }
    }
    println(c.toString())
}