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: Wrong Answer)

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

function sieve(n)
    p = trues(n)
    p[1] = false
    for i in 2:Int(round(sqrt(n))+1)
        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, p)
    i = 1
    res = Vector{Int}[]
    if n >= p[end]
        return [[n, 1]]
    end
    while n > 1
        if i > length(p)
            return [[n, 1]]
        end
        if n % p[i] == 0
            cnt = 0
            while n % p[i] == 0
                n /= p[i]
                cnt += 1
            end
            push!(res, [Int(p[i]), Int(cnt)])
        end
        i += 1
    end
    
    return res
end

function main()
    n = parse(Int, readline())
    
    p = sieve(100000000)
    fc = factor(n, p)

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

    println(ans)
end

main()