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(){
    var n = readLine()!!.split(" ")[0].toLong()
    val sqrt = sqrt(n.toDouble()).toLong()
    val soinsuu = mutableListOf<Long>()
    val kaisuu = mutableListOf<Int>()
    for (i in 2 until sqrt + 1){
        var count = 0
        if (n % i == 0L) {
            while (n % i == 0L) {
                count++
                n /= i
            }
            soinsuu += i
            kaisuu += count
        }
    }
    var ans = 0
    for (i in 0 until kaisuu.size){
        var j = 1
        while (kaisuu[i] >= 0){
            kaisuu[i] -= j
            j++
            ans++
            if (kaisuu[i] < j) break
        }
    }
    if (soinsuu.size == 0 && n != 1L){
        ans = 1
    }
    println(ans)
}