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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define rep(i, n) for(int i=0;i<(int)(n);i++)


long gcd(long a, long b){
    long r, temp;
    if(a<b){
        temp = a;
        a = b;
        b = temp;
    }

    // Euclidean Algorithm 
    r = a%b;
    while(r != 0){
        a = b;
        b = r;
        r = a%b;
    }
    return b;
}

int main(void){
    long a, b;
    long g,count=0;
    long i, j;
    int flag;
    scanf("%ld%ld",&a ,&b);
    g = gcd(a, b);
    for(i=1;i<=g;i++){
        if(g%i == 0){
            flag = 1;
            if(i%j == 0)
                flag = 0;
            for(j=3;j<i;j=j+2){
                if(i%j == 0){
                    flag = 0;
                    break;
                }
            }
            if(flag == 1)   count++;
        }
    }
    printf("%ld\n",count);
    return 0; 
}