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 <string.h>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <set>
#include <map>
#include <vector>

typedef long double ld;
typedef long long ll;

using namespace std;

#define MAXN (1000010)

int p[MAXN];
vector<ll> pp;
ll n;

int main() {
	for (int i = 0; i < MAXN; i++) {
		p[i] = 1;
	}
	for (int i = 2; i < MAXN; i++) {
		if (p[i]) {
			int j = 2 * i;
			while (j < MAXN) {
				p[j] = 0;
				j += i;
			}
		}
	}
	
	
	cin >> n;

	for (int i = 2; i < MAXN; i++) {
		if (p[i]) {
			ll cur = i;
			while (cur <= n) {
				pp.push_back(cur);
				cur *= i;
			}
		}
	}
	sort(pp.begin(), pp.end());
	int ret = 0;
	for (int i = 0; i < pp.size(); i++) {
		if ((n % pp[i]) == 0) {
			n /= pp[i];
			ret++;
		}
	}
	int bp = 1;
	for (int i = 2; i < MAXN; i++) {
		if (!p[i]) {
			continue;
		}
		if ((n % i) == 0) {
			bp = 0;
			break;
		}
	}
	
	if (n > 1) {
		ret += bp;	
	}

	cout << ret;


	return 0;
}