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)

//2019/9/28

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

int euc_alg(int a,int b) {
    int r=a%b;
    while(r!=0) {
        a=b;
        b=r;
        r=a%b;
    }
    return b;
}

int prime(int n) {  /*素数判定機*/
    int num,i;
    if (n==2) {
        return 1;
    }
    else if (n<2 || n%2==0) {
        return 0;
    }
    else {
        num=(int)sqrt(n);
        for (i=3;i<=num;i++) {
            if (n%i==0) {
                return 0;
                break;
            }
        }
        return 1;
    }
}

int main (void) {
    int num1,num2,ans=1,temp,dum,i;
    scanf("%d %d",&num1,&num2);
    if (num2>num1) {
        temp=num1;
        num1=num2;   
        num2=temp;
    }
    temp=euc_alg(num1,num2);
    if (temp==1) {
        printf("%d\n",ans);
    }
    else {
        if (num1%2==0 && num2%2==0) ans++;
        for (i=3;i<=temp;i+=2) {
            if (prime(i)==1) {
                if (num1%i==0 && num2%i==0) ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}