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 <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair <ll, ll> pll;

const int Max = 2e6 + 10;
const int Mod = 1e9 + 7;
const ll Inf = 1LL << 62;

vector <int> prime;
bool isPrime[1000101];

void sieve() {
	int lim = 1000100, sq = sqrt(lim);
	memset(isPrime, true, sizeof isPrime);
	for(int i = 4; i <= lim; i += 2) {
		isPrime[i] = 0;
	}
	for(int i = 3; i <= sq; i += 2) {
		if(isPrime[i]) {
			for(int j = i * i; j <= lim; j += i) {
				isPrime[j] = 0;
			}
		}
	}
	isPrime[1] = 0;
	isPrime[0] = 0;
	for(int i = 1; i < lim; i++) {
		if(isPrime[i]) {
			prime.push_back(i);
		}
	}
}

int main() {
#ifdef Mr_Emrul
	freopen("inputf.in", "r", stdin);
#endif /// Mr_Emrul

	sieve();

	ll n;
	cin >> n;
	bool f = 0;
	ll cnt = 0;
	ll m = n;
	map <ll, ll> mp;
	if(n == 1) {
		cout << 0 << endl;
		return 0;
	}
	for(ll x : prime) {
		if(x > m) {
			break;
		}
		ll now = x;
		ll prv = m;
		while(prv % now == 0 && now <= prv) {
			//cout << ": " << prv << " : " << now << endl;
			prv = m / now;
			now *= x;
			cnt++;
		}
		m = prv;
	}
	if(!cnt)cnt=1;
	cout << cnt << endl;
	return 0;
}