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 <math.h>
int Y[100];
int zhishu_(int x);
signed main()
{
    int A,B,num=0,ans=0; //要自加的向量没有赋初值
    scanf("%d %d",&A,&B);
    for(int i=1; i<=A; i++)
    {
        if(A%i==0&&zhishu_(i)==1)
        {
            Y[num]=i;
            num++;
        }
    }
    for(int j=0; j<num; j++)
    {
        if(B%Y[j]==0)
            ans++;
    }
    printf("%d",ans);
    return 0;
}
int zhishu_(int x)
{
    for(int i=2; i<=sqrt(x); i++)
    {
        if(x%i==0)
            return 0;
    }
    return 1;
}
/*int yueshu(int a)
{
    int Y[100],n=0;
    for(int i=1;i<=a;i++){
            if(a%i==0){
            Y[i-1]=i;
            n++;
            }
    }
    return n;
}*/