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)

from math import floor, sqrt


def func(x):
    return floor((sqrt(1 + 8 * x) - 1) / 2)


N = int(input())
limit = floor(sqrt(N))

primes = [2]

count = 0
flag = True
p = 2
while 1:
    #print(p,N)
    isprime = True
    if N == 1:
        break
    if p > sqrt(N):
        count += 1
        break
    if p > 3:
        for i in primes:
            if i >= sqrt(p):
                break
            if p % i == 0:
                isprime = False
                break
        if not isprime:
            p += 2
            continue
        primes.append(p)
    a = 0
    while N % p == 0:
        #print("while",p, a)
        a += 1
        N = N // p
    count += func(a)
    if p == 2:
        p = 3
    else:
        p += 2
print(count)