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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define N 1000099
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define LL long long
using namespace std;
int n, cnt;
LL a, b;
bool is[N];
int prime[N];
LL gcd(LL x, LL y)
{
    if (y == 0) return x;
    return gcd(y, x % y);
}
void getprime()
{
    is[1] = is[0] = 1;
    prime[0] = 1;
    for (int i = 2; i < N; i++)
    {
        if (!is[i]) prime[++cnt] = i;
        for (int j = 1; j <= cnt && prime[j] * i < N; j++)
        {
            is[prime[j] * i] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}
bool check(LL x)
{
    for (int i = 2; i * i <= x; i++)
    {
        if (x % i == 0) return 0;
    }
    return 1;
}
int cal(LL x)
{
    int res = 0, i = 0;
    for (i = 0; i <= cnt && prime[i] * prime[i] <= x; i++)
    {
        if (x % prime[i] == 0)
        {
            res++;
            if (x / prime[i] != prime[i] && check(x / prime[i])) res++;
        }
    }
    return res;
}
int main()
{
    cin >> a >> b;
    getprime();
    cout << cal(gcd(a, b)) << endl;
    return 0;
}