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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<cstdlib>
#include<iostream>
#include<cmath>
using namespace std;

void Multiplication1() {
	int A, B;
	cin >> A >> B;
	cout << A * B;
}

/*wa*/
void Multiplication2() {
	int N;
	long long int* A;
	long long int ans = 1;
	cin >> N;
	A = new long long int[N];
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < N; i++) {		
		ans *= A[i];
		if (ans < 0) {
			cout << -1;
			return;
		}
	}
	if (ans > (long long int)pow(10, 18)) {
		cout << -1;
		delete[] A;
		return;
	}
	cout << ans;
	delete[] A;
}

void Multiplication3() {
	long long int A, ans;
	long double B;
	cin >> A >> B;
	ans = A * B;
	cout << ans;
}

bool IsPrime(long int num)
{
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false;

	long double sqrtNum = sqrt(num);
	for (long int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0) return false;
	}
	return true;
}

void DivGame() {
	long long int N;
	int e = 1;
	int ans = 0;
	cin >> N;
	if (N == 1) {
		cout << ans;
		return;
	}
	if (IsPrime(N)) {
		cout << 1;
		return;
	}
	for (long long int p = 2; p <= (long double)sqrt(N); p++) {
		if (IsPrime(p)) {
			long int z = p;
			while (N % z == 0) {
				N /= z;
				z *= p;
				ans++;
			}
		}
	}
	cout << ans;
}

int main() {
	//Multiplication2();
	DivGame();
}