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)

n = int(input())
ans = 0
aaa = 2
maxi = 2
def prime(aaa):
    if n != 1:
        i = 2
        while i*i <= aaa:
            if aaa % i == 0:
                return False
            i += 1
        return True
    else:
        return False
def make_divisors(n):
    if prime(n):
        return True
    a = 0
    i = 2
    while i <= n:
        if n % i == 0:
            if a == 0:
                a = i
                n /= i
                i -= 1
                if n == 1:
                    return True
            elif a == i:
                n /= i
                i -= 1
                if n == 1:
                    return True
                pass
            else:
                return False
        i += 1
    return True
def divisors(a, i):
    global ans, aaa, maxi
    while i <= n and i <= a:
        if a % i == 0:
            if make_divisors(i):
                ans += 1
                num = a / i
                aaa = num
                if maxi <= i:
                    maxi = i
                divisors(num, i+1)
                break
        i += 1
if n == 1:
    print(0)
else:
    if prime(n):
        print(1)
    else:
        divisors(n, 2)
        print(ans)