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 Julia (1.4.0) (judge verdict: Accepted)

iscoprime(P, i) = !any(x -> i % x == 0, P)

function sieve(n)
    p = trues(n)
    p[1] = false
    for i in 2:Int(ceil(sqrt(n)))
        if p[i]
            j = 0
            while i*(j+2) <= n
                p[i*(j+2)] = false
                j += 1
            end
        end
    end
    findall(x->x, p)
end

function factor(n)
    i = 1
    res = Tuple{Int, Int}[]
    for a in 2:sqrt(n)
        n%a != 0 && continue
        ex = 0
        while n%a == 0
            ex += 1
            n /= a
        end

        push!(res, (a, ex))
    end
    n != 1 && push!(res, (n, 1))
    
    return res
end

function main()
    n = parse(Int, readline())
    

    fc = factor(n)
    ans = 0
    for (p, c) in fc
        cnt = c
        x = 1
        while x <= cnt
            cnt -= x
            x += 1
            ans += 1
        end
    end

    println(ans)
end

main()