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 Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program d
    use,intrinsic :: iso_fortran_env
    implicit none
    integer(int64)::n,i,j,ans
    integer(int64):: dn(1000000)=0, bn(0:5050)
    logical:: pn(1000000) = .true. 


    read*, n
    pn(1) = .false.

    do i=2,1000000
        if (pn(i)) then
            j=2
            do while(i*j <= 1000000)
                pn(i*j) = .false.
                j=j+1
            end do
        end if
    end do

    do i=2,1000000
        if (pn(i)) then
            do while(mod(n,i) == 0)
                n=n/i
                dn(i) = dn(i)+1
            end do
        end if
    end do
    ans = 0

    if (n > 1) ans=ans+1
    bn(:) = 0
    do i=100,1,-1
        bn((i*(i+1))/2) = i
    end do
    do i=2,5050
        bn(i) = max(bn(i),bn(i-1))
    end do

    do i=1,1000000
        ans=ans+bn(dn(i))
    end do
    print'(i0)', ans

end program d