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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include<math.h>

int isPrime(int x){
    int i;
    if(x == 2) return 2;
    else if(x%2 == 0) return x;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return x;
    return 2;
}

int findFactors(long long int num){
    int i,a,ans=0,b=2,j=2;
    while (num % b == 0){ // 偶数の素数は2のみ
     ans++;
     num /= b;
     b=pow(b,2);
     j++;
    }
    j=2;
    for (i = 3; i <=num; i++){ 
        a=i;// 他の素数で割っていく
        while (num % a == 0){
            ans++;
            num /= a;
            a=pow(a,j);
            j++;
        }
    }
    return ans;
}

int main(void){
    long long int num,nnum,a,ans;
    scanf("%lld",&num);
    if(num==1)printf("%d",0);
    else{
    a=isPrime(num);
    ans=findFactors(num);
    printf("%lld",ans);
    }
    return 0;
}