Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

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

long int gcd(long int x, long int y) {
  if(x <= 0) return y;
  return gcd(y%x, x);
}

int main() {
  long int a, b;
  scanf("%ld%ld", &a, &b);
  long int g = gcd(a, b);
  int ans = 1;
  /*for (int i=1; i<=sqrt((double)g); i++) {
    if(g%i!=0) continue;
    int boo=1;
    for (int j=2; j<=sqrt((double)i); j++) {
      if(i%j!=0) continue;
      boo = 0;
      break;
    }
    if(boo) ans++;

    int l=g/i;
    if (l==i) continue;
    boo=1;
    for (int j=2; j<=sqrt((double)l); j++) {
      if(l%j!=0) continue;
      boo = 0;
      break;
    }
    if(boo) ans++;
  }*/
  for (int i=2; i<=g; i++) {
    if (g%i!=0) continue;
    ans++;
    while (g%i==0) g/=i;
  }
  printf("%d\n", ans);
  return 0;
}