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)

#define _USE_MATH_DEFINES 1
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define MOD 1000000007ll
#define INF 1000000000000000000ll
#define EPS 1e-9
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock

typedef long long         ll;
typedef long double       ld;
typedef pair <ll,ll>      pl;
typedef tuple <ll,ll,ll>  tl;

ll N, A;

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	cin >> N;
	A = 0;
	
	for (ll p = 2; p <= 1000000; p++) {
		ll e = 0;
		while (N % p == 0) {
			N /= p;
			e++;
		}
		ll c;
		for (c = 1; c <= e; c++) e -= c;
		A += c - 1;
	}
	
	cout << A;
	
    return 0;
}