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>
#include <stdlib.h>
#include <math.h>
#include <string.h>

#define ll long long

void swap (int *x, int *y) {
  int temp;

  temp = *x;
  *x = *y;
  *y = temp;

  return;
}


int gcd(int a, int b){
  int c;

  while (b > 0){
    c = a % b;
    a = b;
    b = c;
  }

  return a;
}

int upll(const void*a, const void*b){
  return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;
}

int downll(const void*a, const void*b){
  return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;
}

void sortup(ll*a,int n){
  qsort(a,n,sizeof(ll),upll);
}

void sortdown(ll*a,int n){
  qsort(a,n,sizeof(ll),downll);
}

long int modpow(long int a,long int n,long int mod){
  long int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }
  return ans;
}

long int modinv(long int a,long int mod){
  return modpow(a,mod-2,mod);
}

int max(int a,int b){
  if(a<b){
    return b;
  }
  else{
    return a;
  }
}

int min(int a,int b){
  if(a<b){
    return a;
  }
  else{
    return b;
  }
}

void chmax(int *a,int b){
  if(*a < b){
    *a = b;
  }

  return;
}

void chmin(int *a,int b){
  if(*a > b){
    *a =b;
  }

  return;
}


/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/

int main(void){
  long int n;
  scanf("%ld",&n);
  long int N=n;
  int ans=0;

  for(int i=2;i*i<=N;i++){
    int cnt=0;
    while(n%i==0){
      cnt++;
      n/=i;
    }

    for(int j=1;j<100;j++){
      if(cnt<j*(j+1)/2){
        ans+=j-1;
        break;
      }
    }
  }
  if(n!=1){
    ans++;
  }

  printf("%d\n",ans);

  return 0;
}