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<stdio.h>
#include<math.h>
long long primeFactorization(long long a,int num,int count[10000]){ //素因数分解を出力するプログラム
    long long i,sq;
    if(a%2==0){ //偶数の場合
      while(a%2==0){
        count[num]++;
        a/=2;
      }
      num+=1;
      return primeFactorization(a,num,count); //2で割った値で再帰
    }
    sq = sqrt(a);
    for(i=3;i<=sq;i+=2){ //3以上√a以下の奇数の場合
        if(a%i==0){
          while(a%i==0){
            count[num]++;
            a/=i;
          }
          num+=1;
          return primeFactorization(a,num,count); //割れた値で再帰
        }
    }
    //偶数でも3以上√a以下の奇数の場合でも割り切れない場合
    if(a!=1){ //aが1でないなら、a自身は素数
        count[num]=1;
    }
    return 0;
}
int main(){
    long long a;
    int num=0,sum=0,i,j,count[10000];
    scanf("%lld",&a);
    for(i=0;i<10000;i++)
      count[i]=0;
    primeFactorization(a,num,count);
    for(i=0;i<=num;i++){
      j=1;
      while(count[i]>0){
        sum++;
        count[i]-=j;
        j++;
      }
    }
    printf("%d\n",sum);
    return 0;
}