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>
int main()
{
    //n用来从0输入公约数数组-1了的,q为两两分配后的个数,z为公约数中质数的个数,nz不是质数的个数
    long A,B,n=-1,q=0,G[10000],x=0,z;
    scanf("%ld %ld",&A,&B);
    //公约数放入数组
    for(int i=1;i<=A;i++){
            if(A%i==0&&B%i==0){
                n++;
                G[n]=i;
            }
    }
    //公约数中的质数的个数
    for(int i=1;i<n;i++){
        for(int j=2;j<G[i];j++){
            if(G[1]==2) break;
            if (G[i]%j==0){
                    x++;
                    break;
        }
    }
    }
    z=n-x;
    //组合后的个数
    for(int i=0;i<=z-1;i++){
    for(int j=i+1;j<=z;j++){
            q++;
    }
    }
    printf("%ld",q);
    return 0;
}