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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define nl '\n' 
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long int
#define pii pair <int,int>
ll mod=1000000007;
int main()
{
	fast;
	ll a,b;cin>>a>>b;
	vector<ll>v;
	if(a%2==0 && b%2==0)
	{
		v.push_back(2);
		while(a%2==0)
		{
			a=a/2;
		}
		while(b%2==0)
		{
			b=b/2;
		}

	}
	else
	{
		if(a%2==0)
		{
			a=a/2;
		}
		if(b%2==0)
		{
			b=b/2;
		}
	}
	//cout<<b<<nl;
	ll limit=(sqrt(a),sqrt(b));
	for(int i=3;i<=limit+1;i+=2)
	{
		if(a%i==0 && b%i==0)
		{
			v.push_back(i);
			while(a%i==0)
			{
				a=a/i;
			}
			while(b%i==0)
			{
				b=b/i;
			}
		}
		else
		{
			while(a%i==0)
			{
				a=a/i;
			}
			while(b%i==0)
			{
				b=b/i;
			}
		}
	}
	if(a==b && a>1)
		v.push_back(a);
	
	cout<<v.size()+1;
}