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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.util.*;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		long n = Long.parseLong(sc.next());

		// nが1だったら答えは0で確定
		if (n == 1) {
			System.out.println(0);
			return;
		}

		// nを素因数分解
		HashMap<Long, Long> map = primeFactorize(n);

		// 素因数分解の結果表示
		for (long key : map.keySet()) {
			// System.out.println(key + "^" + map.get(key));
		}

		// 結果を求めて出力
		int result = 0;
		for (long key : map.keySet()) {
			long z = map.get(key);
			for (int j = 1; j < 10000; j++) {
				if (j <= z) {
					z -= j;
					result++;
				} else {
					break;
				}
			}
		}

		System.out.println(result);

	}

	// nを素因数分解して、結果用mapに素因数と指数のペアを入れて返す関数。
	public static HashMap<Long, Long> primeFactorize(long n) {

		long input = n;
		HashMap<Long, Long> map = new HashMap<Long, Long>();

		// nが0以下だったらnullを返す
		if (n <= 0) {
			return null;
		}

		// nが1だったら素因数分解結果は1^1
		if (n == 1) {
			map.put(1L, 1L);
			return map;
		}

		// 素因数分解していく。計算量はO(√n)
		for (long i = 2; i * i <= input; i++) {
			long j = 0;
			while (true) {
				if (n % i == 0) {
					j++;
					map.put(i, j);
					n /= i;
				} else {
					break;
				}
			}
		}

		// 最後に残った数も忘れずに入れる
		if (n != 1) {
			map.put(n, 1L);
		}

		return map;
	}
}