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(args: Array<String>) {
    var n = readLine()!!.toLong()
    val primeExponentialMap = executePrimeFactorization(n)

    var result = 0
    /* TLE
    for ((_, exponential) in primeExponentialMap) {
        var exp = exponential
        var i = 1
        while (i <= exp) {
            result++
            exp -= i
            i++
        }
    }
     */
    primeExponentialMap.forEach{
        var n = it.value
        var i = 1
        while (n >= i) {
            result++
            n -= i
            i++
        }
    }
    println(result)
}

/**
 * 素因数分解し、結果をmap(素数, key素数の指数)として返す
 * @param inputNumber
 * @return map(key=素数, value=key素数の指数)
 */
fun executePrimeFactorization(inputNumber: Long): MutableMap<Long, Long> {
    var n = inputNumber
    val primeExponentialMap = mutableMapOf<Long, Long>()
    // 素数判定なので2から判定していく
    var i = 2L
    // root nまで評価すれば十分 -> i * i <= nで表現
    while (i * i <= n) {
        if (n % i != 0L) {
            continue
        }

        // 指数を数える
        var exponential = 0L
        // 割れるだけ割る
        while (n % i == 0L) {
            exponential++
            n /= i
        }
        // key
        val prime = i
        // map(key=素数, value=key素数の指数)で設定
        primeExponentialMap[prime] = exponential

        i++
    }

    // 1でなければ最後の素数が残っているのでmapに足す
    if (n != 1L) {
        primeExponentialMap[n] = 1L
    }

    return primeExponentialMap
}