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)

import math
n = int(input())
ans = [int(i) for i in range(n+1)]
ans[1] = 0
i = 2
sosu = []
while i <= int(math.sqrt(n)):
    if ans[i] != 0:
        j = i
        sosu.append(i)
        for j in range(j,(n//j)+1):
            ans[i*j] = 0
    i += 1
f = []
tmp = n
for i in range(len(sosu)):
    if tmp%sosu[i]==0:
        cnt = 0
        while tmp%sosu[i]==0:
            cnt += 1
            tmp = tmp//sosu[i]
        f.append([sosu[i],cnt])
if tmp != 1:
    f.append([tmp,1])
count = 0
for i in range(len(f)):
    j = 1
    while j <= f[i][1]:
        n = n//(f[i][0]*j)
        f[i][1] -= j
        count += 1
        j += 1
#print(f)
print(count)