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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict";

const main = arg => {
    const inputList = arg.trim().split("\n");
    // 回答を記述

    // 回答を記述
    let N = parseInt(inputList[0]);
    const SqrtN = Math.sqrt(N);
    let numList = [] // 素因数分解した後の素数の累乗リスト
    // Nを素因数分解する
    for (let i = 2; i <= SqrtN; i++) {
        let sum = 0;
        while (true) {
            if (N % i !== 0) {
                break;
            }
            N /= i;
            sum++;
        }
        numList.push(sum);
    }

    // √Nまで割って1より大きい=かなり大きい素数のため、素数**1と判定でき、1を追加する
    if (N !== 1) {
        numList.push(1);
    }

    if (numList.length === 0) {
        console.log(0);
    } else {
        let result = 0;
        numList.filter(ele => ele !== 0).forEach(ele => {
            for (let i = 1; i <= ele; i++) {
                ele -= i;
                if (ele < 0) {
                    break;
                }
                result++;
            }
        })

        console.log(result);
    }

}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));