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: Time Limit Exceeded)

#include<stdio.h>
int hantei(int y);

int hantei(int y){
    int h[100];
    int i,cnt,x;
    cnt=0;
    x=1;
    for(i=2;y>1;){
        if(y%i == 0){
            y=y/i;
            h[cnt]=i;
            cnt++;
            i = 2;
         }else{
             i++;
         }
    }
    for(cnt = cnt - 1;cnt!=0;cnt--){
        if(h[cnt] != h[cnt-1]){
            x=0;
            break;
        }
    }
    return x;
}
int main(void){
    long long int N,i;
    int z,count;
    count = 0;
    scanf("%lld",&N);
    for(i=2;i<=N;i++){
        z=0;
        if(N%i == 0) z = hantei(i);
        if(z != 0) {
            N = N/i;
            count++;
        }
    }
    printf("%d",count);
}