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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pii pair<int, int>
#define pll pair<long long , long long>
#define pb push_back
#define ll long long
#define ld long double
#define precision(x,d) cout<<fixed<<setprecision(d)<<x
#define fill(a,b) memset((a),(b),sizeof((a)))
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);srand(time(NULL));
ll power(ll x,ll y,ll m){ll r=1;for(;y;y>>=1){if(y&1)r=r*x%m;x=x*x%m;}return r;}
const ll mod = 1e9 + 7;
const ld pi = 3.14159265358979;
    
int run_test(){
	ll a, b;
	cin >> a >> b;
	ll g = __gcd(a, b);
	//now I have to find all the prime factors of g
	int count = 0;
	for(int i=2; i<=sqrt(g); i++){
		if(g % i == 0){
			count++;
			while(g % i == 0){
				g /= i;
			}
			if(g == 1) break;
		}
	}
	if(g != 1) count++;
	cout << count + 1;
	return 0;
}
    
int main(){
   FAST;
    ll t;
    t = 1;
     
    while(t--){
        run_test();
    }
}