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)

fun main() = Thread(null, ::abc169d, "(^o^)", 128 * 1024 * 1024).apply { start() }.join()

@OptIn(ExperimentalStdlibApi::class)
fun abc169d() {
    val n = readLine()!!.toLong()

    var currN = n
    var answer = 0

    for (div in 2..Math.sqrt(n.toDouble()).toLong()) {
        if (!isPrime(div)) continue
        var z = div
        while (z <= n) {
            if (currN % z == 0L) {
                currN /= z
                answer++
            }
            z *= div
        }
    }

    if (n > 1 && isPrime(n) && currN % n == 0L) answer++

    println(answer)
}

private fun isPrime(n: Long): Boolean {
    val sqrt = Math.sqrt(n.toDouble()).toLong()
    for (i in 2..sqrt) if (n % i == 0L) return false
    return true
}