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;
bool is_prime(ll x){
    for(int i=2;i*i<=x;i++){
    	if(x%i==0)return 0;
	}
	return 1;
}
vector<ll> v1;
ll dives(ll n){
	ll ans=0;
    for(int i=1;i*i<=n;i++){
    	if(n%i==0){
    		ans++;
    		if(n/i!=i)v1.push_back(n/i);
    		v1.push_back(i);
		}
	}
}

void solve(){
	ll a,b,ans=0,tmp;
	scanf("%lld %lld",&a,&b);
	if(a>b)swap(a,b);
	dives(a);
	for(int i=0;i<v1.size();i++){
		if(b%v1[i]==0&&is_prime(v1[i])){
			ans++;
		}
	}
	printf("%lld",ans);
}

signed main(){
    int t=1;
    //scanf("%d",&t);
    for(int i=1;i<=t;i++){
        solve();
    }
    return 0;
}