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>

void swap(long *a, long *b){
  long tp;
  tp=*a;
  *a=*b;
  *b=tp;
  return;
}

int isprime(long n){
  long i;
//  if(n<=2)return 1;
  for(i=2;i*i<=n;i++){
    if(n%i==0) return 0;
  }
  return 1;
}

int main()
{
  long a,b,newa,newb,i;
  int ans;
  scanf("%ld%ld", &a,&b);
  if(a<b) swap(&a,&b);
  while(newb=a%b){
    a=b;
    b=newb;
  }
  //lcd=b;
  ans=1;
  if(b%2==0){
    ans++;
    while(b%2==0) b/=2;
  }
  if(b%3==0){
    ans++;
    while(b%3==0) b/=3;
  }
  for(i=5;i<=b/2;i+=2) {
    if(b%i==0 && isprime(i)) ans++;
  }
  printf("%d\n",ans);
  return 0;
}