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>
#include<math.h>

typedef unsigned long long int lui;

lui i = 0, j = 0;

lui Euclid(lui A, lui B){
    lui p = 0, q = 0, qa = 0;
    if(A > B){
        q = A%B;
        p = B;
        qa = q;
        while(q != 0){
            q = p%q;
            p = qa;
            qa = q;
        }
    }else if(A < B){
        p = B;
        q = B%A;
        qa = q;
        while(q != 0){
            q = p%q;
            p = qa;
            qa = q;
        }
    }else{
        p = A;
    }

    return p;
}

int main(void){
    lui A = 0, B = 0;
    lui p;
    lui *cdv;
    lui k = 0;
    lui cnt = 0;
    int flg = 1;
    

    scanf("%llu %llu", &A, &B);

    p = Euclid(A, B);

    for(i = 1; i <= p; i++){
        if(p%i == 0){
            for(j = 2; j <= sqrt(i); j++){
                if(i%j == 0){
                    flg = 0;
                    break;
                }
            }
            if(flg == 1){
                cnt++;
            }
            flg = 1;
        }
    }

    printf("%llu", cnt);

    free(cdv);
    return 0;
}