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<cmath>
#include <bits/stdc++.h>
using ll = long long;
using namespace std;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
#define mins(x,y) x=min(x,y)
#define maxs(x,y) x=max(x,y)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;
static const int INF = 1000000007;
#define mp make_pair
#define sz(x) int(x.sise())
#define mod 1000000007
#define reps(i,s,n) for(int i = s; i < n; i++)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
deque<int> deq;
#define fi first
#define se second
//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
typedef pair<int, int> P;
int l[1000005];
int r[45];
int main(){
    ll n;
    cin >> n;
    ll N=n;
    while(n%2==0){
        l[2]++;
        n/=2;
    }
    for(int i=3;i<sqrt(N)+1;i+=2){
        if(n%i==0){
            l[i]++;
            n/=i;
        }
    }
    ll ans=0;
    if(N!=1&&n==N){
        ans++;
    }
    r[1]=1;
    for(int i=2;i<44;i++){
        r[i]+=r[i-1]+i;
    }
    int k=0;
    while(r[k+1]<=l[2]){
        k++;
    }
    ans+=k;
    //cerr<<l[3]<<endl;
    for(int i=3;i<sqrt(N)+1;i+=2){
        int m=0;
        while(r[m+1]<=l[i]){
            m++;
        }
        ans+=m;
        //cerr<<ans<<endl;
    }
    cout<<ans<<endl;
    return 0;
}