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>

unsigned long gcd(unsigned long a,unsigned long b)
{
   //printf("a=%ld,b=%ld\n",a,b);
	if (a%b==0)
	{
		return(b);
    }
	else
	{
		return(gcd(b,a%b));
    }
}

int main(void){
    // Your code here!
    unsigned long A,B,t;
    scanf("%lu %lu\n",&A,&B);
    unsigned long GCD;
    if(A<B){
        t=B;
        B=A;
        A=t;
    }
    GCD = gcd(A,B);
    //sqrt_GCD = sqrt(GCD);
   // printf("GCD:%lu\n",GCD);
   
    unsigned long int i,ct,tmp;
    ct = 1;
    tmp = GCD;
    if(tmp%2==0){
            ct++;
            while(1){
                tmp=tmp/2;
                if(tmp%2!=0) break;
            }
    }
    if(tmp!=1){
    for(i=3;GCD>=i;i=i+2){
        if(tmp%i==0){
            ct++;
            while(1){
                tmp=tmp/i;
                if(tmp%i!=0) break;
            }
        }
        //printf("%ld:%ld\n",i,tmp);
        if(tmp==1) break;
    }
    }
    
    printf("%lu\n",ct);
    
    
    return 0;
}