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>

long long int gcd(long long int x, long long int y)
{
  while(x > 0 && y > 0) {
    if(x > y) {
      x = x % y;
    }
    else {
      y = y % x;
    }
  }

  if(x == 0) {
    return y;
  }
  else {
    return x;
  }
}

long long int approximate_sqrt(long long int x)
{
  long long int mean;
  long long int l = 0;
  long long int r = x;

  if(x == 1) {
    return 1;
  }
  
  while(1) {
    mean = (l + r) / 2;
    if(mean * mean <= x && x < (mean+1) * (mean+1)) {
      break;
    }
    else if(mean * mean < x) {
      l = mean;
    }
    else {
      r = mean;
    }
  }

  return mean;
}

int number_of_factors(long long int x)
{
  long long int root_x = approximate_sqrt(x);
  long long int x_iter = x;
  int num = 0;

  if(x == 1) {
    return 0;
  }

  if(x_iter % 2 == 0) {
    num++;
  }
  while(x_iter % 2 == 0) {
    x_iter = x_iter / 2;
  }

  long long int i = 3;
  while(i <= root_x) {

    if(x_iter % i == 0) {
      num++;

      do {
        x_iter = x_iter / i;
      } while(x_iter % i == 0);
    }

    i += 2;
  }

  if(x_iter > 1) {
    num++;
  }

  return num;
}

int main(void)
{
  long long int a, b;
  scanf("%lld%lld", &a, &b);

  printf("%d\n", 1 + number_of_factors(gcd(a,b)));
  return 0;
}