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>
using namespace std;
 
#define int long long int
#define vi vector<int>
#define vvi vector<vi>
#define read(a)  for(int i = 0; i < n; i++) cin >> a[i];
#define print(a)  for(int i = 0; i < n; i++) cout << a[i] << " ";
#define pb push_back
#define pql priority_queue<int>
#define pqs priority_queue<int, vi, greater<int>>
#define pqlv priority_queue<vi>
#define pqsv priority_queue<vi, vvi, greater<vi>>
// #define endl '\n'
#define N 10000002
#define MOD 1000000007

int bs(int n) {
    int low = 0, high = n, ans = 1;
    while(low <= high) {
        int mid = low + (high - low) / 2;
        if((mid * (mid + 1) / 2) <= n) {
            ans = mid;
            low = mid + 1;
        }
        else {
            high = mid - 1;
        }
    }
    return ans;
}

signed main() {
    int n;
    int ans = 0;
    for(int i = 2; i * i <= n; i++) {
        int cnt = 0;
        while(n % i == 0) {
            n /= i;
            cnt++;
        }
        ans += bs(cnt);    
    }
    cout << ans << endl;
}