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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll gcd(ll a,ll b){
	if(a<b)swap(a,b);
	if(a%b==0)return b;
	else return gcd(b,a%b);
}
void solve(){
	ll a,b,tmp;
	int ans=0;
	scanf("%lld %lld",&a,&b);
	tmp=gcd(a,b);
	for(int i=2;i*i<=tmp;i++){
		if(tmp%i==0){
			ans++;
			while(tmp%i==0)tmp/=i;
		}
	}
	if(tmp>1)ans++;
	printf("%d",ans+1);
}
signed main(){
    int t=1;
    //scanf("%d",&t);
    for(int i=1;i<=t;i++){
        solve();
    }
    return 0;
}