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>
#include<time.h>
#define SIZE 1000


typedef unsigned long long int lui;

lui Euclid(lui A, lui B){
    if(B == 0){
        return A;
    }else{
        return Euclid(B, A%B);
    }
}

lui mod_p(lui U, lui a, lui p){
    return (U*U + a)%p;
}

lui rho_f(lui p){
    lui d = 0;
    srand((unsigned)time(NULL));
    lui U = rand()%p;
    lui V = U;
    lui a = rand()%(p-2);
    int i = 0;
    while(1){
        U = mod_p(U, a, p);
        V = mod_p(mod_p(V, a, p), a, p);
        d = Euclid(abs(U-V), p);
        if(d != 1 && d != p){
            return p/d;
        }else if(d == p){
            srand(++i);
            U = rand()%p;
            V = rand()%p;
            a = rand()%(p-2);
        }
    }
}

lui prime_n(lui p){
    lui i = 0;
    for(i = 2; i < sqrt((double)p); i++){
        if(p%i == 0){
            return 0;
        }
    }
    return 1;
}

int main(void){
    lui A = 0, B = 0;
    lui p;
    lui d = 1;
    lui cnt = 1;
    lui i = 1, j = 0;
    int flg = 1;

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

    p = Euclid(A, B);

    while(prime_n(p) == 0){
        p = rho_f(p);
        cnt++;
    }

    if(p != 1){
        cnt++;
    }

    printf("%llu", cnt);

    return 0;
}