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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define Rep(i,l,r)for(int i=(int)l;i<(int)r;i++)
#define rep(i,n)Rep(i,0,n)
typedef long long lld;
typedef long double Lf;
#define Ma(x,y) ((x)>(y)?(x):(y))
#define mi(x,y) ((x)<(y)?(x):(y))
#define N 1000001//絶対
int z = 0, large=0; lld ans = 0;

int fac(lld n, int a[]) {
	for (lld k = 2; k <= n; k++) {
		int o = sqrt(n);
		if (n % k == 0) { a[k]++; n /= k; k = 1; z++; }
		if (k > o && !z) { z = -1; large++; break; }
	}
	return z;
}

int main() {
	lld n;
	scanf("%lld", &n);
	if (n > 1) {
		int a[N] = { 0 };
		if (fac(n, a) == -1) { ans++; }
		else {
			int p = sqrt(n);
			rep(i, p+1) {
				if (a[i]) {
					int b[11] = { 1,3,6,10,15,21,28,36,45,55,66 };
					rep(j, 11) {
						if (a[i] < b[j]) { ans += j; break; }
					}
				}
			}
		}
	}
	if (large)ans++;
	printf("%lld", ans);
}