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 kotlin.math.sqrt

fun main(){
    val N = readLine()!!.toLong()

    if(N == 1L){
        println(0)
        return
    }

    var x = N
    val pSet = mutableSetOf<Long>()
    for(i in 2..sqrt(N.toDouble()).toInt()){
        while(x % i == 0L){
            pSet.add(i.toLong())
            x /= i
        }
    }

    if(pSet.isEmpty()){
        println(1)
        return
    }

    var ans = 0
    x = N
    pSet.forEach { p ->
        var pp = p
        while (x % pp == 0L){
            ans++
            x /= pp
            pp *= p
        }
    }
    println(ans)
}