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>

int main() {
   int N;
   scanf("%d",&N);
   int count = 0;
   int cnt = 0;
   int a[100];
   int ac = 0;
   if(N == 1) {
      printf("%d\n",0);
   } else {
      for(int j = 1; j <= N; j++) {
         if(N % j == 0) {
            cnt++;
         }
      }
      if(cnt == 2) {
         printf("%d\n", 1);
      } else {
         int cnt = 0;
         for(int p = N; p >= N; p--) {
           if(N % p == 0) {
            cnt++;
           }
           if(cnt == 2) {
              int z = p;
              for(int i = 1; N == 1; i++) {
                 if(N % z == 0) {
                    N = N / z;
                    a[ac] = z;
                    ac++;
                    count++;
                 } else {
                    z = z*p;
                 }
              }
           }
         }
         printf("%d\n", count);
      }
   }
}