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(){
    val num = readLine()!!.toLong()

    var i = 2L
    var x = num
    var count = 0L
//    var y = num
    val insuuList = mutableListOf<Long>()
//    val answerList = mutableListOf<Int>()
    while (i <= x + 1) {
        var jousannyou = 1L
        while (insuuList.size > 0 && x % (jousannyou * insuuList.last())  == 0L){
            val nijou = jousannyou * insuuList.last()
            x /= nijou
            jousannyou = nijou
            count++
        }
        if (x % i == 0L) {
            insuuList.add(i)
//            x /= i
//            count++
        }
            i++
    }
//    insuuList.forEach {
//        if (y % it == 0){
//            y /= it
//            answerList.add(it)
//        }
//    }
    if (num < 2) println(0)
    else println(count)
}