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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

#d
n = int(input())
import math
def is_prime(n):
    if n == 1: return False
    for k in range(2, int(math.sqrt(n)) + 1):
        if n % k == 0:
            return False
    return True
if is_prime(n):
    print("1")
    
else:
    used = []
    ans = 0
    for i in range(2,10**12):
        if is_prime(i):
            if n%i ==0:
                n = n/i
                used.append(i)
                ans += 1    

        else:
            for j in used:
                k = j**2
                count = 2
                flag = 0

                while k <= i:
                    k = j**count
                    if k == i:
                        if n % i == 0:
                            n = n/i
                            ans += 1
                            flag = 1
                            break
                    count+= 1
                if flag == 1:
                    break
        if i > n:
            break
    print(ans)