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: Accepted)

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

//#pragma gcc optimize("O3,unroll-loop")
//#pragma gcc target("sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,abm,mmx,popcnt,tune=native")
#define int long long
//#define double long double

int prime[1000001];
vector<int> p;

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    for (int i = 2; i <= 1000000; i++)
        prime[i] = 1;
    for (int i = 2; i <= 1000000; i++)
    {
        if (prime[i] == 1)
            p.push_back(i);
        for (int j : p)
        {
            if (i * j > 1000000)
                break;
            prime[i * j] = j;
            if (i % j == 0)
                break;
        }
    }

    int n;
    cin >> n;

    int ans = 0;
    for (int i : p)
    {
        int cnt = 0;
        while (n % i == 0)
            cnt++, n /= i;

        int t = 1;
        while (cnt >= t)
            ans++, cnt -= t, t++;
    }
    cout << ans + (n != 1) << '\n';

    return 0;
}

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