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: Wrong Answer)

#include"stdio.h"
#include"stdlib.h"
int gcd( int A, int B){
  if( A>B ){
    return gcd( A-B, B);
  }
  else if( B>A ){
    return gcd( A, B-A);
  }
  else{
    return A;
  }
}
int* divisor(int num){
  int i,j=0;
  int* list = (int*) calloc(1000,sizeof(int));
  for(i=1;i<=num;i++){
    if(num%i == 0){
      list[j++] = i;
    }
  }
  return list;
}
int main(void){
  int A,B,C,i,j,count,ct=0;
  int flag=1;
  int* list;
  scanf("%d %d",&A,&B);
  C = gcd( A, B);
  
  // output'C' is max gcd(A,B)
  //printf("C:%d\n",C);

  // list is sosuu
  list = divisor(C);
  for(i=0; list[i]!=0;i++){
    //printf("list[%d]:%d ", i, list[i]);
  }
  count=i;
  //printf("count:%d\n",count);

  for( i=0; i<count; i++){
    for( j=2; j<list[i]; j++){
      if( list[i]%j==0 ){
        flag=0;
        break;
      }
    }
    if( flag==1 ){
      //printf("list[%d]:%d\n", i, list[i]);
      ct++;
    }
    flag=1;
  }
  
  printf("%d\n", ct);
  free(list);
  return 0;
}