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<iostream>
#include<cmath>
#include<math.h>
#include<vector>

long long gcd(long long x, long long y){
	if(x<y){
		long long tmp= x;
		x=y;
		y=tmp;
	}

	if(x%y==0)
		return y;
	else
		return gcd(y, x%y);
}

int main(){
	long long n, m;
	std::cin>>n>>m;

	long long gcdNM= gcd(n, m);
	long long sqGcd= sqrtl(gcdNM);

	std::vector<bool> isPrime(sqGcd+1, true);
	isPrime[0]= false;
	isPrime[1]= false;

	for(int i=2; i<isPrime.size(); i++){
		if(!isPrime[i])
			continue;

		for(int j=2*i; j<isPrime.size(); j+=i)
			isPrime[j]= false;
	}

	std::vector<long long> primeList;
	for(int i=0; i<isPrime.size(); i++)
		if(isPrime[i])
			primeList.push_back((long long)i);

	std::vector<bool> includePrime(primeList.size(), false);

	for(int i=0; i<primeList.size();){
		if(gcdNM% primeList[i] != 0){
			i++;
			continue;
		}

		includePrime[i]= true;
		gcdNM= gcdNM/primeList[i];
	}

	long long ans=1;

	for(int i=0; i<primeList.size(); i++)
		if(includePrime[i])
			ans++;

	if (gcdNM > sqGcd)
		ans++;

	std::cout<<ans<<std::endl;

	return 0;
}