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)

import java.io.*
import java.util.*
import kotlin.math.*

fun main(){
	val f = BufferedReader(InputStreamReader(System.`in`))

	var n = f.readLine().toLong()

	val triangle = LongArray(46)
	triangle[0] = 0L
	for(i in 1..45){
		triangle[i] = triangle[i-1]+i.toLong()
	}



	var answer = 0L
	var i = 2L
	while(i*i <= n){
		var count = 0L
		while(n%i == 0L){
			n/=i
			count++
		}
		if(count == 0L){
			i++
			continue
		}
		var l = 1
		var r = 40
		var ans = -1
		while(l <= r){
			var mid = l + (r-l)/2
			//println("$i $mid")
			if(triangle[mid] <= count){
				ans = mid
				l = mid+1
			} else {
				r = mid-1
			}

		}
		//println("$i $ans")
		i++
		answer += ans.toLong()
	}

	if(n != 1L) answer++
	println(answer)
}