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>

#define LIM 1000000
int sieve[LIM+1], primes[LIM+1], pcnt = 0;

int main(void) {
	long long int n;
	scanf("%lld", &n);

	if (n == 1) {
		puts("0");
		return 0;
	}

	sieve[0] = sieve[1] = 1;
	primes[pcnt++] = 2;
	for (int i = 3; (long long int) i*i <= n; i += 2) {
		if (sieve[i])
			continue;
		primes[pcnt++] = i;
		for (long long int j = (long long int) i*i; j <= LIM; j += i)
			sieve[j] = 1;
	}

	int prime = 1, ans = 0;
	for (int i = 0; i < pcnt; i++) {
		int cnt = 0;
		while (n % primes[i] == 0) {
			cnt++;
			n /= primes[i];
		}
		if (!cnt)
			continue;

		prime = 0;
		for (int j = 1; j <= cnt; j++) {
			cnt -= j;
			ans++;
		}
	}

	if (prime)
		ans++;
	printf("%d\n", ans);
}