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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int table[1000];
int bisect_right(int x,int a[],int left,int right){
    int mid;
    right++;
    while (left < right) {
        mid = (left + right) / 2;
        if (a[mid] <= x) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
}


int counter(int i) {
    if (i == 0) {
        return 0;
    }
    return bisect_right(i,table,0,999) - 1;
    
}

void solve(long long N, int *ans ){
    int cnt = 0;
    while (N % 2 == 0) {
        cnt++;
        N /= 2;
    }
    *ans += counter(cnt);
    int now = 3;
    cnt = 0;
    while (now * now <= N) {
        while (N % now == 0) {
            cnt++;
            N /= now;
        }
        *ans += counter(cnt);
        cnt = 0;
        now += 2;
    }
    if (N != 1) {
        *ans += 1;
    }
}



int main(void){
    long long N;
    scanf("%lld", &N);
    for (int i = 0; i < 100; i++) {
        table[i] = i * (i + 1) / 2;
    }
    int ans = 0;
    solve(N, &ans);
    printf("%d\n",ans);
    return 0;
}