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

int gcd(int a, int b){
    if (a < b) {
        int tmp;
        tmp = a;
        a = b;
        b = tmp;
    }
    if (b == 0) {
        return a;
    }
    return gcd(a % b, b);
}

int main(void){
    int A,B;
    scanf("%d %d", &A, &B);
    int cnt = 1;
    int g = gcd(A , B);
    int i = 2;
    while (g > 1) {
        if (i * i > g) {
            printf("%d\n", g == 1 ? cnt:cnt + 1 );
            return 0;
        }
        if (g % i == 0) {
            do {
                g /= i;
            } while (g % i);
            cnt++;
        } else {
            i++;
        }
    }
    printf("%d\n", g == 1 ? cnt:cnt + 1 );
}