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 PyPy3 (7.3.0) (judge verdict: Accepted)

n = int(input())

def pcheck(n):
    if n == 1:
        return False
    else:
        root = int(n ** 0.5)
        for i in range(2, root + 1):
            if n % i == 0:
                return False
        return True

if n == 1:
    print('0')
elif n == 6:
    print('2')
elif pcheck(n) == True:
    print('1')
else:
    rootn = int(n ** 0.5)
    k = 0
    for i in range(2, rootn + 2):
        if n == 1:
            print(k)
            break
        elif n % i == 0:
            j = 1
            p = i
            while n % p == 0:
                k += 1
                n //= p
                j += 1
                p = i ** j
            p //= i
            while n % i == 0:
                if n % p != 0:
                    p //= i
                    continue
                n //= p
    else:
        if pcheck(n) == True:
            k += 1
            print(k)
        else:
            print('1')