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>

int main()
{
    long long a = 0, b = 0, min = 0, i = 0;
    int k = 0, count = 0;
    int *p = NULL;
    scanf("%I64d %I64d", &a, &b);
    min= a < b ? a:b;
    p = (int *)calloc(min, sizeof(int));
    for(i = 3; i <= min; i += 2)
    {
        if (a % i == 0 && b % i== 0)
        {
            for (k = 0; k < count; k++)
            {
                if (i % p[k] == 0) break;
            }
            if (k == count)
            {
                count++;
                p[k] = i;
            }
        }
    }
    if (a % 2 == 0 && b % 2 == 0) count++;
    count++;
    printf("%d", count);
    free(p);
    return 0;
}