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 (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

/****************/
/*  素因数分解  */
/****************/
#include <stdio.h>
#include <stdlib.h>

int main( )
{
    char  c[128];
    long int   n0,n;   // 整数
    long int   ns=2;   // 素因数
    long int   j;      // 乗数
    long int k;
    long int ans = 0;

    //printf("\n★素因数分解★\n");
    //printf("\n分解する整数: "); fflush(stdout);
    gets(c); n0=atoll(c);                    // 整数を入力

    for( n=n0,ns=2; n>=ns; ns++ )
      {
        for( j=0; n%ns==0; j++ )            // 割り切れる間繰り返す
         {
           n/=ns;
         }

        if ( j==0 ) continue;               // 1回も割り切れなかった

        //printf("素因数:%d  乗数:%d\n",ns,j);

        for(k=1; k<100; k++){
            if(j - k >= 0){
                ans++;
                j -= k;
            }
        }
     }
     printf("%d\n", ans);
  return 0;
}