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: Accepted)

fun main() {

    /**
     * 素因数分解
     * 試し割り法
     */
    fun factorize(number: Long): List<Pair<Long, Int>> {

        var n = number

        val res = mutableListOf<Pair<Long, Int>>()

        for (i in 2..Math.sqrt(n.toDouble()).toLong() + 1) {
            //割り切れなかったら素因数ではない
            if (n % i != 0L) {
                continue
            }
            //割り切れた数だけ++
            var count = 0
            while (n % i == 0L) {
                n /= i
                count++
            }
            res.add(i to count)
        }

        // √n以上の値で素数があったら
        if (n != 1L) {
            res.add(n to 1)
        }
        return res
    }

    val n = readLine()!!.toLong()

    val list = factorize(n)
    var result = 0
    for (item in list) {
        var (_, count) = item
        for (i in 1..count) {
            if (0 <= count - i) {
                count -= i
                result++
            }
        }
    }
    println(result)

}