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<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5
#define PI 3.141592653589793238

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
void cin(ll *n){ scanf("%lld",&(*n)); }
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
double POW_d(double a, double b){double c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod_MOD1(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
ll mod_p(ll n ,ll p){n+= n<0?((-n)/p+1)*p:0; return n%=p;}
ll change_into_num(char s[] , ll len, ll p){ return !p ? 0 : POW(10,p-1)*(s[len-p]-'0') + change_into_num(s,len,p-1); }

void lr_lower( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am < val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am <= val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
void lr_upper( int *l, int *r, ll am, ll val , int type ){ (type<3) ? ( am <= val ?  ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ) : ( am < val ? ( *l = (*l+*r)/2 ) : ( *r= (*l+*r)/2 ) ); }
int cmp_lower( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a>=b ? 1 : 0 ) : ( a>b ? 1 : 0 ) ; }
int cmp_upper( ll a, ll b, int type ){ return (type==1) ? ( a==b ? 1 : 0 ) : (type==2) ? ( a<=b ? 1 : 0 ) : ( a<b ? 1 : 0 ) ; }
// return smallest p  which meets  a[p]==val :1  >=:2   >:3
ll lower_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_lower(&l,&r,a[ (l+r)/2 ],val,type);  return cmp_lower(a[l],val,type) ? l : cmp_lower(a[r],val,type) ? r : -1;  }
// return biggest p  which meets   a[p]==val :1  <=:2   <:3
ll upper_bound( ll a[], int l, int r, ll val , int type ){  while(r-l>1) lr_upper(&l,&r,a[ (l+r)/2 ],val,type); return cmp_upper(a[r],val,type) ? r : cmp_upper(a[l],val,type) ? l : -1; }
// count i  which meets ai==x
ll count(ll a[], int l, int r, ll x){  int p = lower_bound(a,l,r,x,1);  return p==-1 ? 0 : upper_bound(a,p,r,x,1)-p+1; }

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;}
int cmp_string( const void * a , const void * b ) {  return strcmp( (char *)a , (char *)b ); }  // qsort((void*)s,n,sizeof(s[0]),int_sort );
int cmp_char(const void * a, const void * b) { return *(char *)a - *(char *)b;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
void sort_string(int n,int size,char s[][size]){ qsort( (void*)s , n , sizeof(s[0]) , cmp_string ); }
void sort_char(char *s){ qsort( (void *)s , strlen(s) , sizeof(char) , cmp_char ); }
ll unique_string(ll n ,ll size, char s[][size]){ ll ans=1; rep(i,1,n) if( strcmp(s[i],s[i-1]) ) ans++; return ans; }
ll unique_num(ll n , ll a[]){ ll ans=1; rep(i,1,n) if( a[i]!=a[i-1] ) ans++; return ans; }

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}

/*nが素数かどうか判定*/
ll judge_prime(ll n){
    if(n==2) return 1;
    if(n==1 || n%2==0) return 0;
    ll a=sqrt(n); if(a%2==0) a--;
    repp(i,3,a+1,2){if(n%i==0) return 0;}
    return 1;
}


int main(void){
    // fgets(s,sizeof(s),stdin); 
    // char ;
    ll n;
    ll ans=0;
    cin(&n);
    // scanf("%s",);
    // ll [n];
    if(n%2==0){
        ll tmp=2;
        while(n%tmp==0){
            n/=2;
            ans++;
            tmp*=2;
        }
    }
    repp(i,3,POW(10,6)+1,2){
        if(n%i==0){
            if(judge_prime(i)){
                ll tmp=i;
                while(n%tmp==0){
                    n/=tmp;
                    ans++;
                    tmp*=i;
                }
            }     
        }
    }
    if(n>POW(10,6)){
        if(judge_prime(n)) ans++;
    }

    printf("%lld\n",ans);
    return 0;
}