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: Accepted)

#include<stdio.h>

int isnp[1001001] = {1,1};
int ecnt[1001001];

int main(void){
    register long i,j;
    for(i = 2;i < 1001001;i++)if(!isnp[i])for(j = 2;i*j < 1001001;j++)isnp[i*j] = 1;
    long long n;
    scanf("%lld",&n);
    if(n == 1){
        puts("0");
        return 0;
    }
    static int nispf;
    for(i = 2;i < 1001001;i++)if(!isnp[i])while(!(n % i)){n /= i;ecnt[i]++;}
    if(n != 1)nispf = 1;
    static long long ans;
    for(i = 2;i < 1001001;i++)if(ecnt[i]){
        j = 1;
        do{ecnt[i] -= j;ans++;}while(ecnt[i] >= ++j);
    }
    if(nispf)ans++;
    printf("%lld",ans);
}