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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
long long a,b,c,ans;
struct node{
	int order;
	int i;
}aa[100001];
int qpow(int x,int y)
{
	int tmp=1;
	while(y){
		if(y&1)tmp*=x;
		y>>=1;
		x*=x;
	}return tmp;
}
int cmp(node x,node y)
{
	return x.order<y.order;
}
bool isprime(long long x)
{
	if(x<=1)return 0;
	for(long long i=2;i<=sqrt(x);i++)
	{
		if(x%i==0)return 0;
	}return 1;
}
long long gcd(long long x,long long y){return y==0?x:gcd(y,x%y);}
void fenjie(long long c){
	long long tmp=c;
	for(long long i=2;i<=sqrt(c);i++)
	{
		if(tmp%i==0){
		//	cout<<i<<endl;
			ans++;
			while(tmp%i==0)tmp/=i;
		}
	}
	if(isprime(tmp))ans++;
}
int main()
{
	cin>>a>>b;
	c=gcd(a,b);
	fenjie(c);
	cout<<ans+1;
	return 0;
}