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: Runtime Error)

#include"bits/stdc++.h"

#define rep(i, N) for(int i = 0;i < N;i++)
typedef long long ll;
const int mod = 1e9 + 7;

using namespace std;

//エラトステネスの篩、1からnまでの素数を探す
class Sieve {
private:
	vector<bool> isprime;
public:
	Sieve(long long n) {
		isprime = vector<bool>(n + 1, true);
		isprime[0] = false;
		isprime[1] = false;

		for (int i = 2; i * i <= n; i++) {
			if (isprime[i])for (int j = i + i; j <= n; j += i)isprime[j] = false;
		}
	}

	bool operator()(ll n) { return isprime[n]; }
};


int main(void) {
	ll N; cin >> N;
	Sieve sieve(N);

	ll ans = 0;

	for (int i = 2; i <= N; i++) {
		if (sieve(i)) {
			ll x = i;
			while(true) {
				if (N % x == 0) {
					N /= x;
					ans++;
					x *= i;
				}
				else {
					break;
				}
			}
		}
	}

	cout << ans << endl;

	return 0;
}