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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii  = pair<int,int>;
const int INF=1001001001;
#define rep(i,n) for(int i=0; i<(n); i++)
#define rep2(i,x,n) for(int i=x; i<(n); i++)
#define MOD 1000000007
#define EPS 10e-8
int gcd(int a,int b){return b?gcd(b,a%b):a;}    //最大公約数
int lcm(int a, int b) {return a * b / gcd(a, b);}   //最小公倍数
const ll MAXN = 1001024; // b-a < MAXN, sqrt(b) < MAXN

using namespace std;
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
ll A,B;
//エラトステネスの篩
bool is_prime[ 10000 + 1 ];
vector<int> P;
void Eratosthenes( const int N )
{
    for( int i = 0; i <= N; i++ )
    {
        is_prime[ i ] = true;//初期化
    }    
    for( int i = 2; i <= N; i++ )
    {
        if( is_prime[ i ] )
        {
            for( int j = 2 * i; j <= N; j += i )
            {
                is_prime[ j ] = false;
            }
            P.emplace_back( i );
        }
    }
}
int main()
{
    cin >> A>>B;
    ll ans = 1;
    if(A>=B){
        Eratosthenes(B);
    }
    else
    {
        Eratosthenes(A);
    }
    if(A>=B){
    for( const auto &x: P )
    {
        if(A%x==0)
            ans++;
    }
    }
    else{
        for( const auto &x: P )
    {
        if(B%x==0)
            ans++;
    }
    }
    cout << ans << endl;
    return 0;
}