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: Accepted)

#include<bits/stdc++.h>
#define rep(i,n,m) for(int i = (n); i <(m); i++)
#define rrep(i,n,m) for(int i = (n) - 1; i >=(m); i--)
using namespace std;
using ll = long long;
const ll MAX_N = 10000100;

vector<int> is_prime;
void init_prime()
{
    is_prime.assign(MAX_N, 1);
    is_prime[0] = 0;
    for (ll i = 2; i*i < MAX_N; ++i)
    if (is_prime[i])
    {
        for (ll j = i*i; j < MAX_N; j += i)
            is_prime[j] = 0;
    }
}

int main()
{
    init_prime();
    ll a, b;
    cin >> a >> b;

    int cnt = 1;
    for (ll n = 2; n <= MAX_N; ++n)
    if (is_prime[n] and a % n == 0 and b % n == 0)
        ++cnt;
    
    ll g = __gcd(a, b);
    // cout << g << endl;
    rep(i, 2, MAX_N)
    if(is_prime[i])
        while(g % i == 0)
            g /= (ll)i;
    
    if (g > 1) ++cnt;

    cout << cnt << endl;
    return 0;
}