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 Ruby (2.7.1) (judge verdict: Wrong Answer)

N = gets.chomp.to_i


# 素数で割っていってカウントする
# ルートNまでの素数をまず列挙する
# それで順番に悪のを繰り返す
# それをハッシュにいれておく
# 最後に足す
# 2が3回とかだったらまた別の計算方法をする？


# Nが素数ならそれで割って終了
# 割れないならルートNまで割っていく
# 累乗数でないならスキップ
if N == 1
  puts 0
  exit
end

ans = 0

num = N

2.upto(Math.sqrt(N).to_i) do |i|
  if num % i == 0
    num /= i
    ans += 1
  end
end

if num > Math.sqrt(N)
  ans += 1
end

if num == N
  puts 1
  exit
end

puts ans