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<stdlib.h>
#include<math.h>
#include<string.h>

int W_Length(char *str){
	int length = 0;
	while(*str++ != '\0')
		length++;
	return length;
}

double deg(double theta){//deg to theta
return theta*(2*3.14159265359)/360;
}

int cmp( const void *p, const void *q ) {
    return -*(int*)p + *(int*)q;
}
//qsort(a,n,sizeof(int),cmp);
//char s[n][11]

int cmp_s(const void * left, const void * right)
{
	char *left_char = (char *)left;
  char *right_char = (char *)right;
  return strcmp(left_char, right_char);
}
//qsort(s,n,sizeof(s[0]),cmp_s);

int end(void){
return 0;
}

long long int waru(long long int *n,long long int x){
  long long int y=x,j=0;
  while(1){
    if((*n)%y==0){
    *n=(*n)/y;
      y=x*y;
      j++;
    }
    else{
      break;
    }
  }

    return j;  

}

int main (void){

long long int n,count=0,i,j,k,temp;
scanf("%lld", &n);
  temp=n;
  if(n%2==0){
  count+=waru(&n,2);
  }

    	for(i=3;i<sqrt(temp);i+=2)
	{
		k=0;
		for(j=3;j<=sqrt(i);j+=2)
		{
			if(i%j==0)
			{
				k=1;
              break;
			}
        }
		if(k==0&&n%i==0){
        count+=waru(&n,i);
        
        }
	}
  
printf("%lld\n",count);
return 0;
}