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: Wrong Answer)

#D - Div Game
N = int(input())

def Prime_factorize(n):
    cnt = 0
    A = []
    #2の倍数で割っていく
    z = 2
    while n%z==0 and (n>=z):
        cnt += 1
        n //= z
        A.append(z)
        #割るたびに2倍
        z *= 2
    #割る始めの数をfとして保持
    while n%2 == 0:
        n//=2
    f = 3
    z = 3
    #sqrt(n)以内に全ての素因数は含まれる
    while f*f <= N:
        if n%z == 0:
            cnt += 1
            n //= z
            A.append(z)
            #割るたびにf倍
            z = z*f
        #fで割り切れなかったら(f+2)してf_divもリセット
        else:
            f += 2
            z = f
    #残ったnが以前に選んだもので割り切れないとき
    if n != 1 and all((n%i != 0 for i in A)):
        A.append(z)
        cnt += 1
    #素数のとき
    if cnt == 0:
        cnt = 1
    return cnt
if N == 1:
    ans = 0
else:
    ans = Prime_factorize(N)
print(ans)