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 <math.h>

void swap(long long *a, long long *b)
{
    *a = *a ^ *b;
    *b = *a ^ *b;
    *a = *a ^ *b;
    return;
}

int main()
{
    long long a = 0, b = 0;
    long i = 0, max = 0;
    int count = 1;
    int flag1 = 0, flag2 = 0, flag = 0;
    scanf("%lld %lld", &a, &b);
    if (a > b) swap(&a, &b);
    max = (long)sqrt(a)+1;
    if (b % a == 0 && a>1) flag = 1;
    while (a % 2 == 0)
    {
        a /= 2;
        flag1 = 1;
    }
    while (b % 2 == 0)
    {
        b /= 2;
        flag2 = 1;
    }
    if (flag1==1 && flag2==1) count++;
    for (i = 3; i <= max; i += 2)
    {
        if (i > a || i > b) break;
        flag1 = 0;
        flag2 = 0;
        while (a % i == 0)
        {
            a /= i;
            flag1 = 1;
        }
        while (b % i == 0)
        {
            b /= i;
            flag2 = 1;
        }
        if (flag1==1 && flag2==1) count++;
    }
    if (flag==1 && count==1) count++;
    printf("%d", count);
    return 0;
}