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 <bits/stdc++.h>

int mem[45][45];
int64_t V;

int dp(int N, int v) {
  if (N == 0) return 0;
  if (mem[N][v] != -1) return mem[N][v];
  int best = -2;
  for (int i = v + 1; i <= N; ++i) {
    best = std::max(dp(N - i, i) + 1, best);
  }
  mem[N][v] = best;
  return best;
}

std::unordered_map<int, int> m;

int main() {
  memset(mem, -1, sizeof(mem));
  scanf("%lld", &V);

  int64_t i = 2;
  int count = 0;
  int64_t origV = V;
  while (V != 1 && i <= int(sqrt(origV) + 1)) {
    if (V % i == 0) {
      m[i]++;
      V /= i;
    } else i++;
  }
  if (i > int(sqrt(origV) + 1)) {
    printf("1\n");
    return 0;
  }
  for (auto kv : m) {
    // printf("prime: %d %d %d\n", kv.first, kv.second, dp(kv.second, 0));
    count += dp(kv.second, 0);
  }
  printf("%d\n", count);
}