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>

void check(int n, int ds[], int *cnt) {
    for (int i=0; i<(*cnt); i++) {
        if (n % ds[i] == 0) return;
    }
    ds[(*cnt)++] = n;
    return;
}



int main() {
    int A,B;
    scanf("%d %d", &A, &B);
    int ds[66000000];
    int cnt = 0;
    int min = (A<B)?A:B;
    for (int i=2; i<=min; i++) {
        if ((A%i == 0) && (B%i == 0)) {
            check(i, ds, &cnt);
        }
    }
    printf("%d\n", cnt + 1);
    return 0;
}