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;
typedef long long ll;
#define endl 	"\n"

typedef vector< ll > 	vi;
typedef vector< vi > 	vvi;
typedef pair< ll,ll > 	ii;
#define pb 		push_back
#define all(c) 		(c).begin(),(c).end()
#define tr(c,i) 	for(typeof((c).begin() i = (c).begin(); i != (c).end(); i++)
#define present(c,x) 	((c).find(x) != (c).end())

#define MOD 	1000000007
#define INF 	(int)1e9
#define MEM(a, b) memset(a, (b), sizeof(a))

std::vector<ll> prime(10000000,1);
void fun(ll n)
{
	prime[0]=0; prime[1]=0;

	for(ll i=2; i*i<=n;i++)
	{
		if(prime[i])
		{
			for(ll j=i*i; j<=n; j+=i)
				prime[j]=0;
		}
	}
}

int main()
{	 ios_base::sync_with_stdio(false);
	 cin.tie(NULL);
	

	ll a,b;
	cin>>a>>b;
	ll k=min(a,b), cnt=1;
	fun(k);
	for(ll j=2; j*j<= k; j++)
	{
		if(a%j==0 && b%j==0 && prime[j]  )
			cnt++;
	}
	cout<<cnt;
	return 0;
}