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 main() {
    long long int n;
    scanf("%lld", &n);
	if (n == 1) {
		printf("0");
		return 0;
	}
	int count2 = 0;
    int ans=0;
	while (n % 2 == 0) {
		//printf("2 ");
		n /= 2;
		count2++;
	}
    int aa=1;
    while(count2>=1){
        //printf("--%d\n",count2);
        count2-=aa;
        aa++;
        if(count2>=0){ans++;}
    }
    //printf("--%d\n",ans);
    int count;
	for (int i = 3; i <= 1000000; i++) { 
	    count=0;
		while (n % i == 0) {
			//printf("%d ", i);
			n /= i;
			count++;
		}
		aa=1;
		
		while(count>=1){
		//printf("--%d\n",count);
        count-=aa;
        aa++;
        if(count>=0){ans++;}
        //printf("--ans %d\n",ans);
		}
		count=0;
	}
	if (n > 100000) {
		ans++;
	}
    printf("%d", ans);
	return 0;
}