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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;
long long int gcd(long long int x,long long int y){
	if(max(x,y)%min(x,y)==0){
		return min(x,y);
	}else{
		return gcd(max(x,y)%min(x,y),min(x,y));
	}
}
int main()
{
	long long int a,b;
	cin>>a>>b;
	vector<long long int> n;
	long long int g=gcd(a,b);
	n.push_back(1);
	long long int gcd=g;
	
	for(long long int i=2;i<=gcd;i++){
		if(g%i==0){
			n.push_back(i);
			while(g%i==0){
				g=g/i;
			}
		}
	}
	cout<<n.size()<<endl;
}