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)

n = int(input())
from collections import Counter
#a,b = map(int,input().split())
#L = list(map(int,input().split()))
def pr(n):
    L = []
    for i in range(2,int(n**0.5)+1):
        while n % i==0:
            L.append(i)
            n = n//i
    if n != 1:
        L.append(n)
    return L
dic = {1:1,2:1,3:2,4:2,5:2,6:3,7:3,8:3,9:3,10:4,11:4,12:4,13:4,14:4,15:5,16:5,17:5,18:5,19:5,20:5,21:6,22:6,23:6,24:6,25:6,27:6,28:7,29:7,30:7,31:7,32:7,33:7,34:7,35:7,36:8,39:8,40:8}
L = pr(n)
#print(L)
L = Counter(L)
num = 0
for k,v in L.items():
    num +=dic[v]
print(num)