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)

from collections import Counter
from itertools import combinations
import numpy as np

# 引用元
# https://note.nkmk.me/python-prime-factorization/
def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

def get_list(counter, index):
    ans = []
    for key, val in sorted(counter.items()):
        if val > index:
            ans += [key] * index
        else:
            ans += [key] * val
    return ans

def choice_min_data(counter, min_num):
    temp = sum(counter.values())
    temp_min_num = 10**12
    co_li = None
    for index in range(1, temp+1):
        flag = False
        to_comb_list = get_list(counter, index)
        for co in combinations(to_comb_list, index):
            check_val = np.prod(co)
            if temp_min_num > check_val > min_num:
                co_li = co
                temp_min_num = check_val
                flag = True
                break

        # 最小更新せず = 終了
        if co_li is not None and flag == False:
            break

    if co_li is None:
        return None, -1
    else:
        # print(co_li, temp_min_num)
        return counter - Counter(co_li), temp_min_num

def main():
    # ルール
    # 素数p, 整数e: z == p ** e
    # N / z = int(x)
    # zは被らない
    # N を N/zに置き換える
    N = int(input())

    if N <= 1:
        print(0)
        return

    counter = Counter(prime_factorize(N))
    # print(counter)
    ans = 0
    min_num = 0
    while counter is not None:
        counter, min_num = choice_min_data(counter, min_num)
        if min_num == -1:
            break
        ans += 1
    
    print(ans)

if __name__ == "__main__":
    main()