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 x = num
    var count = 0
    val sosuuList = decom(num)
    sosuuList.forEach {
        var kisuu = 1L
        while (x % (it * kisuu) == 0L){
            val keisango = it * kisuu
            x /= keisango
            count++
            kisuu = keisango
        }
    }
    if (isPrime(x)) count++

    if (num < 2) println(0)
    else println(count)
}



// 範囲内の素数を取得
// fromだけ指定すると戻り値の個数で素数判定ができる
fun prime(from : Long, to : Long = from) : List<Long> {
    return (from..to).filter { i -> isPrime(i) }
}

// 素因数分解
fun decom(value : Long) : List<Long>{
//    if (value == 1L) return listOf(1)
    if (value == 1L) return emptyList()
    val max = Math.sqrt(value.toDouble()).toLong()
    return prime(2, max).filter { value % it == 0L }
}

fun isPrime(num: Long): Boolean {
    if (num < 2)
        return false
    else if (num == 2L)
        return true
    else if (num % 2 == 0L) return false // 偶数はあらかじめ除く

    val sqrtNum = Math.sqrt(num.toDouble())
    var i = 3
    while (i <= sqrtNum) {
        if (num % i == 0L) {
            // 素数ではない
            return false
        }
        i += 2
    }

    // 素数である
    return true
}