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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
#include<math.h>
#include<limits.h>
#define MAX 1000004
typedef long long ll;
int main(){
    ll N;
    ll prime_index=0;
    ll prime_seq[MAX]={0};
    bool prime_flag[MAX];    //素数ならtrue;
    for(ll i=2;i<MAX;i++) prime_flag[i]=true;
    scanf("%lld",&N);
    //素数列作るよ
    for(ll i=2;i<MAX;i++){
        if(!prime_flag[i]) continue;
        for(ll j=2;i*j<MAX;j++)
            prime_flag[i*j]=false;
    }
    for(ll i=2;i<MAX;i++)
        if(prime_flag[i]) prime_seq[prime_index++]=i;
    //素数列作り終わったよ
    prime_index--;
    for(int i=0;i<prime_index;i++){
        if(N%prime_seq[i]!=0){
            break;
        }
    }
    ll ans=0;
    //素因数分解始まるよ
    ll prime_fac[MAX]={0};
    bool f=false;
    while(N!=1){
        ll i=0;
        while(N%prime_seq[i]!=0 && prime_index>i)i++;
        if(prime_index<=i){
            ans++;
            break;
        }
        N/=prime_seq[i];
        prime_fac[prime_seq[i]]++;
        f=true;
    }
    //素因数分解終わったよ

    //総和のテーブル作るよ
    ll table[100];
    table[1]=1;
    for(ll i=2;i<=100;i++)
        table[i]=table[i-1]+i;
    //総和のテーブル作り終わったよ
    for(ll i=0;i<prime_index;i++){
        if(!prime_fac[prime_seq[i]]) continue;
        ll j=1;
        while(table[j]<=prime_fac[prime_seq[i]]) j++;
        j--;
        ans+=j;
    }
    printf("%lld",ans);
    return 0;
}