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: Accepted)

#include<cmath>
#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>

inline void read(int &T) {
	int x=0;bool f=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=!f;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	T=f?-x:x;
}

long long n,ans;

long long qpow(int a,int b) {
	long long ans=1,base=a;
	while(b) {
		if(b&1) ans*=base;
		base*=base;
		b>>=1;
	}
	return ans;
}

void branch(long long a) {
	for(int i=2;1ll*i*i<=a;++i) {
		int flag=0;
		long long base=qpow(i,flag+1);
		while(a%base==0) {
			a/=base;
			++flag;
			base=qpow(i,flag+1);
			++ans;
		}
		while(a%i==0) a/=i;
	}
	if(a!=1) ++ans;
}

int main() {
	std::cin>>n;
	branch(n);
	std::cout<<ans<<'\n';
	return 0;
}