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 <math.h>
using namespace std;
#define int long long
int a,b;
int ans=1;
int gcd(int a,int b){
	if (a<b) swap(a,b);
	if (b==0) return a;
	else return gcd(b,a%b);
}

signed main()

{
	ios::sync_with_stdio(false);
	cin>>a>>b;
	int c=gcd(a,b);
	int oc=c;
	for (int i=2;i<=sqrt(oc);i++){
		if (c%i==0&&i<=c){
			ans++;
			//cout<<i<<endl;
			while (c%i==0) c/=i;
		}
	}

	for (int i=sqrt(oc);i>=1;i--){
		int ni=oc/i;
		if (c%ni==0){
			if (ni<=1) continue;
			ans++;
			//cout<<ni<<endl;
			while (c%ni==0&&ni<=c) c/=ni;
		}
	}
	cout<<ans<<endl;
	return 0;
}