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>

int main()
{
    int a, b, count = 1, temp, i, p[100000], x, y, ans = 0;
    scanf("%d %d", &a, &b);
    if (a > b)
    {
        temp = a;
        a = b;
        b = temp;
    }
    for (i = 1; i <= a; i++)
    {
        if (a % i == 0 && b % i == 0)
        {
            p[count] = i;
            count = count + 1;
        }
    }
    for (i = 1; i < count; i++)
    {
        if (i != count - 1)
        {
        }
    }
    for (i = 1; i < count; i++)
    {
        x = p[i];
        y = 2;
        while (1)
        {

            if (y * y > x)
            {
                ans = ans + 1;
                break;
            }
            if (x % y == 0)
            {
                break;
            }
            y = y + 1;
        }
    }
    printf("%d", ans);
    return 0;
}