Problem:
Score : 400 points

Problem Statement

Given is a positive integer N. Consider repeatedly applying the operation below on N:

First, choose a positive integer z satisfying all of the conditions below:

z can be represented as z=p^e, where p is a prime number and e is a positive integer;

z divides N;

z is different from all integers chosen in previous operations.

Then, replace N with N/z.

Find the maximum number of times the operation can be applied.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum number of times the operation can be applied.

Sample Input 1

24

Sample Output 1

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=12.)

Choose z=3 (=3^1). (Now we have N=4.)

Choose z=4 (=2^2). (Now we have N=1.)

Sample Input 2

1

Sample Output 2

0

We cannot apply the operation at all.

Sample Input 3

64

Sample Output 3

3

We can apply the operation three times by, for example, making the following choices:

Choose z=2 (=2^1). (Now we have N=32.)

Choose z=4 (=2^2). (Now we have N=8.)

Choose z=8 (=2^3). (Now we have N=1.)

Sample Input 4

1000000007

Sample Output 4

1

We can apply the operation once by, for example, making the following choice:

z=1000000007 (=1000000007^1). (Now we have N=1.)

Sample Input 5

997764507000

Sample Output 5

7

Solution in Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextLong()
    println(problem169d(n))
}

fun problem169d(n: Long): Int {
    if (n == 1L) return 0

    /**
     * 素数判定
     */
    fun isPrime(num: Int): Boolean {
        if (num < 2) return false
        if (num == 2) return true
        if (num % 2 == 0) return false

        val sqrtNum = Math.sqrt(num.toDouble()).toInt()
        for (i in 3..sqrtNum step 2) {
            if (num % i == 0) return false
        }
        return true
    }

    /**
     * エラトステネスのふるい。1からnまでの素数を列挙する
     */
    fun sieveOfEratosthenes(n: Int): List<Int> {
        val primes = (2..n).toMutableList()
        for (i in 2..n) {
            if (isPrime(i)) {
                for (j in i * 2..n step i) {
                    primes.remove(j)
                }
            }
        }
        return primes
    }

    val sosu = sieveOfEratosthenes(10000)
    val sosu2 = mutableListOf<Long>()
    for (i in 0 until sosu.size) {
        for (j in 1 until 10L) {
            sosu2.add(sosu[i] * j)
        }
    }
    val sorted = sosu2.sorted()
    val used = mutableSetOf<Long>()
    var count = 0
    var n2 = n
    while (true) {
        val prev = n2
        loop@ for (i in 0 until sorted.size) {
            val so = sorted[i]
            if (used.contains(so)) continue
            if (n2 % so != 0L) {
                continue
            }
            n2 /= so
            used.add(so)
            break@loop
        }
        if (prev == n2) break
        count++
    }

    return Math.max(count, 1)
}