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: Time Limit Exceeded)

fun main() {
    var n = readLine()!!.toLong()

    val counter = mutableMapOf<Long, Int>()
    while (n % 2L == 0L) {
        counter[2L] = (counter[2L] ?: 0) + 1
        n /= 2L
    }

    var factor = 3L
    val factorList = mutableListOf(2L, 3L)
    while (n >= factor) {
        while (n % factor == 0L) {
            counter[factor] = (counter[factor] ?: 0) + 1
            n /= factor
        }

        factor += 2L
        while (factorList.any { factor % it == 0L }) {
            factor += 2L
        }
        factorList.add(factor)
    }

    val max = counter.map { it.value }.max() ?: 0
    val triangular = (1..max).map {
        it * (it + 1) / 2
    }

    val result = counter.map { (_, value) ->
        val index = triangular.indexOfLast { value >= it }
        if (index == -1) {
            0
        } else {
            index + 1
        }
    }.sum()

    println(result)
}