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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <math.h>
using namespace std;

int main(void){
    long long A, B, N, x, i, t, p[20], c, d;
    scanf("%lld%lld", &A, &B);
    if (A>B) {
        t=A;
        A=B;
        B=t;
    }
    N=(long long)(floor(sqrt(A)));
    c=0;
    for (i=2; i<=N; i++) {
        d=0;
        while (1) {
            if (A%i==0) {
                if (d==0) {
                    c++;
                    p[c]=i;
                    d=1;
                    // printf("!(%lld, %lld)\n", A, i);
                }
                A/=i;
                // printf("?(%lld, %lld)\n", A, i);
            }
            else break;
            // printf("(%lld, %lld)\n", i, A%i);
        }
    }
    if (c==0) {
        c=1;
        p[1]=A;
    }
    // for (i=1; i<=c; i++) printf("%lld\n", p[i]);
    x=0;
    for (i=1; i<=c; i++) {
        if (B%p[i]==0) x++;
    }
    printf("%lld", x+1);
    return 0;
}