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>

#define FOR(i, l, r) for(ll i = l; i < r; i++)
#define rep(i, N) FOR(i, 0, N)

#define MOD 1000000007
#define INF 1000000000

using ll = long long int;
using namespace std;

typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;

int main() {
    ll n, ntmp; 
    cin >> n;
    map<ll,ll> ret;
    vector<ll> nsm((ll)sqrt(n)+2, 0);
    nsm[0] = 0;
    FOR(i,1,(ll)sqrt(n)+2) nsm[i] = nsm[i-1]+i;
    FOR(i,0,10) cout << nsm[i];

    ntmp = n;
    bool flag = false;
    for(ll i = 2; i * i <= ntmp; i++) {
        while(ntmp % i == 0) {
            flag = true;
            ret[i]++;
            ntmp /= i;
        }
    }
    if(ntmp != 1) ret[ntmp] = 1;

    ll ans = 0;
    auto itr = ret.begin();
    while(itr != ret.end()){
        ll tmp = itr->second;
        ans += (upper_bound(nsm.begin(),nsm.end(),tmp)-nsm.begin())-1;
        itr++;
    }

    cout << ans << "\n";
    return 0;
}