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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

#pragma GCC diagnostic push
#pragma GCC diagnostic ignored "-Wunused-result"

ll A, B, cd; bool isPrime[1000002];ll ans = 0;
int main(){
    memset(isPrime, 1, sizeof isPrime);
    scanf("%lld %lld", &A, &B);
    cd = __gcd(A, B);
    isPrime[0] = isPrime[1] = false;
    for (ll i = 2; i*i <= 1000000; i++) {
        if (isPrime[i]) {
            if(cd%i ==0) ans++;
            for (ll j = i * i; j <= 1000000; j += i)
                isPrime[j] = false;
        }
    }
    ans++;
    printf("%lld\n", ans);
    return 0;
}
#pragma GCC diagnostic pop