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>
typedef long long int ll;
using namespace std;


int c(int x)
{
    int m = 0, t = 0, ret = 0;
    while(m <= x)
    {
        ret++;
        t++;
        m+=t;
    }
    return ret-1;
}

void solve()
{
    ll n;
    cin >> n;
    ll ogn = n;
    ll cnt = 0, twos = 0;

    while(n%2==0)
    {
        twos++;
        n/=2;
    }
    cnt += c(twos);

    for(ll i = 3; i*i<=ogn; i+=2)
    {
        twos = 0;
        while(n%i==0)
        {
            twos++;
            n/=i;
        }
        cnt += c(twos);
    }
    if(n==ogn)
        cnt++;
    cout << cnt << endl;
}


int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll t;
    solve();
    return 0;
}