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: Accepted)

import math

def append_factor(f, p):
    if p not in f:
        f[p] = 0
    f[p] += 1

def append_factors(n, f, p):
    num_factors = 0
    while n % p == 0:
        num_factors += 1
        n //= p
    if num_factors != 0:
        f[p] = num_factors
    return n

def factors(n):
    l = int(math.sqrt(n)) + 1
    ret = {}
    n = append_factors(n, ret, 2)
    n = append_factors(n, ret, 3)
    i = 6
    while i*i <= n + 1:
        p1, p2 = i - 1, i + 1
        n = append_factors(n, ret, i - 1)
        n = append_factors(n, ret, i + 1)
        i += 6
    if n != 1:
        append_factor(ret, n)

    return ret

def minimum_sum(x):
    # find the minimum integer n where n(n+1)/2 <= x
    # n^2 < n(n+1) <= 2x < (n+1)(n+2) < (n+2)^2

    n = int(math.sqrt(x))
    while (n + 1)*(n + 2) <= 2*x:
        n += 1

    return n

def main():
    N = int(input())
    fs = factors(N)
    # print(fs)
    ans = 0
    for f in fs:
        ans += minimum_sum(fs[f])
    print(ans)

if __name__ == '__main__':
    main()