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: Accepted)

import java.util.*

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

fun problem169d(n: Long): Long {
    /**
     * 素因数分解
     */
    fun computePrimeFactorList(n: Long): List<Pair<Long, Long>> {
        val p = mutableListOf<Pair<Long, Long>>()
        var primeFactor = n
        for (i in 2..(Math.sqrt(n.toDouble()) + 1).toLong()) {
            var count = 0L
            while (primeFactor % i == 0L) {
                primeFactor /= i
                count++
                if (primeFactor == 1L) break
            }
            p.add(i to count)
        }
        if (primeFactor != 1L) p.add(primeFactor to 1L)
        return p
    }

    val pf = computePrimeFactorList(n)
    var count = 0L
    for (i in 0 until pf.size) {
        var (a, b) = pf[i]
        var j = 1
        while (b > 0) {
            if (b < j) break
            b -= j
            count++
            j++
        }
    }
    return count
}