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> 

int main() {
  long int A, B, a, b;
  long int r;
  int counter = 1;
  //  int rtmp=0;
  
  scanf("%ld %ld", &A, &B);

  a = A; b = B; 
  
  if(A < B){
    a = B; b = A;
  }  

  do{
    r = a%b;
    a = b;
    b = r;
  }while(r!=0); 

  if(a==1){
    printf("1\n");
    return 0;
  }

  //  printf("GCD %d\n",a);
  
  if(a%2 == 0){
    //    printf("D %d\n",2);
    counter++;
    a /= 2;
    while(a%2 == 0){
      a /= 2; 
    }
  }
  
  for(r=3;r<a/2;r += 2){
    if(a%r == 0){
      //      printf("D %d\n",r);
      counter++;
      a /= r;
      while(a%r == 0){
	a /= r; 
      }
    }
  }
	
  if(a>=r) counter++;
  
  printf("%d\n", counter); 
  
  return 0;
}