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 C# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Collections.Generic;
using System.Threading.Tasks;

class Program
{
	//Main
	static void Main(string[] args)
	{
		//読み込む
		long N = long.Parse(Console.ReadLine());

		long ans = 0;
		long div = 2;

		long limit = (long)Math.Sqrt(N);

		while(N >= div){
			if(N % div == 0){
				N /= div;
				ans ++;

				//debug
				//Console.WriteLine(div);
			}

			div ++;

			//ルートまで行ってもペアが見つからない場合、それは素数である
			//ちょうど二乗の場合を含んではならない
			if(div >= limit){
				if(ans == 0){
					ans ++;
				}
				break;
			}
		}

		//現在の余りも素数の場合、加える必要あり
		if(N > 1 && N+1 != div){
			long lim2 = (long)Math.Sqrt(N);
			long cnt = 0;
			long div2 = 2;

			while(N >= div2){
				if(N % div2 == 0){
					N /= div2;
					cnt ++;
					break;
				}

				div2 ++;

				//ルートまで行ってもペアが見つからない場合、それは素数である
				//ちょうど二乗の場合を含んではならない
				if(div2 >= lim2){
					break;
				}
			}

			if(cnt == 0){
				ans ++;
			}
		}

		//出力
		Console.WriteLine(ans);
	}
}