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.

Examples

Input

24


Output

3


Input

1


Output

0


Input

64


Output

3


Input

1000000007


Output

1


Input

997764507000


Output

7
import java.util.Scanner;

public class Main {

	static long bignum = 1000000000+7;

	public static void main(String[] args){

		Scanner sc = new Scanner(System.in);

		long n = sc.nextLong();

		int[] soin = new int[1000010];
		for(int i =0;i<soin.length;i++) {
			soin[i]=0;
		}
		long target = 2;
		int folder = 0;
		boolean bit = false;
		boolean sosu  =true;
		long nn = n;
		long upper = (long)Math.sqrt(n)+3;
		while(target<n+1) {
			if(n%target==0) {
				n = n/target;
				soin[folder]++;
				sosu =false;
				bit  =true;
			}else {
				target++;
				if(bit) {
					folder++;
					bit =false;
				}
				if(target>upper) {
					if(!sosu) {
						soin[folder]++;
					}
					break;
				}
			}
		}
		int counter =0;
		for(int i=0;i<soin.length;i++) {
			int checker =1;
			while(checker<=soin[i]) {
				counter++;
				soin[i]-=checker;
				checker++;
			}
		}
		if(sosu&&(nn!=1)) {
			System.out.println(1);
		}else {
			System.out.println(counter);
		}
	}



	static long divn(int target, int times) {//10_9+7のあまり累乗版
		int sqtimes = (int)Math.sqrt(times);
		long ans = 1;
		for(int i =0;i<sqtimes;i++) {
			ans = (ans * target)%(1000000000+7);
		}
		long holder = ans;
		ans = 1;
		for(int i =0;i<sqtimes;i++) {
			ans = (ans*holder)%(1000000000+7);
		}
		for(int i=0;i<times - sqtimes*sqtimes;i++) {
			ans = (ans * target)%(1000000000+7);
		}
		return ans;
	}

	static long div(long a) {//10_9+7のあまり
		if(a<0)a+=1000000000+7;
		return a%(1000000000+7);
	}

}
/*
class City{
	int order;
	int prefecture;
	int year;
	int number;
	City(int order,int prefecture, int year){
		this.order = order;
		this.prefecture = prefecture;
		this.year = year;
	}
}
*/
