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)

//Your limitation—it’s only your imagination
#include <bits/stdc++.h>

#define full(x,n) x,x+n+1
#define full(x) x.begin(),x.end()
#define finish return 0

#define putb push_back
#define f first
#define s second

#define putf push_front
#define gainb pop_back

#define gainf pop_front
#define len(x) x.size()

//Dream it. Wish it. Do it

// 1/b%mod=b^(m-2)%mod

typedef double db;
typedef long long ll;

//Never forget health

const ll ary=1e5+5;
const ll mod=1e9+7;
const ll inf=1e18;

using namespace std;
ll a,b;
map<ll,ll> mp;
vector<ll> q;
int main(){
	cin>>a>>b;
	for(int i=1;i<=sqrt(a);i++){
		if(a%i==0){
			mp[i]=1;
			if(i*i!=a){
				mp[a/i]=1;
			}
		}
	}
	for(int i=1;i<=sqrt(b);i++){
		if(b%i==0){
			if(mp[i]){
				q.putb(i);
			}
			if(i*i!=b){
				if(mp[b/i]){
					q.putb(b/i);
				}
			}
		}
	}
	sort(full(q));
	ll lcm=1,ans=1;	
	for(int i=1;i<len(q);i++){
		if(__gcd(lcm,q[i])==1){
			ans++;
			lcm=lcm*q[i]/__gcd(q[i],lcm);
		}		
	}
	cout<<ans;
}
//Push yourself, because no one else is going to do it for you