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 <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mod 1000000007
#define ll long long
#define F first
#define S second
#define pii pair<double,int>
#define pi atan(1)*4
#define BS(x,xx) binary_search(x.begin(),x.end(),xx)
#define ALL(x) x.begin(),x.end()
#define int ll
main()
{
	int a,b;
	cin>>a>>b;
	int q=__gcd(a,b);
	int re=1;
	for(int i=2;i*i<=q;i++)
	{
		if(q%i==0) re++;
		while(q%i==0)
		{
			q/=i;
		}
	}
	if(q!=1) re++;
	cout<<re<<endl;
}