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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,cc,n) for(int i=cc;i<=n;++i)
#define drep(i,cc,n) for(int i=cc;i>=n;--i)
#define sz(s) (int)(s.size())
#define vecprint(v) rep(i,0,v.size()-1)cout << v[i] << " ";cout << endl;
#define mod 1000000007
using namespace std;

unsigned long long gcd(unsigned long long a,unsigned long long b)
{
	if (a%b==0){
		return(b);
    }
	else{
		return(gcd(b,a%b));
    }
}

int main(){
    unsigned long long a,b,ans,n,k;

    cin >> a >> b;

    if(a<b)swap(a,b);
    k = gcd(a,b);

    ans = 1;

    n=2;
    if(k%n==0){
        ans++;
        while(k%n==0){
            k /= n;
        }
    }
    ++n;
    while(k >= n){
        if(n%3==0 && n/3 >1)continue;
        if(n%5==0 && n/5 >1)continue;
        if(n%7==0 && n/7 >1)continue;
        if(n%11==0 && n/11 >1)continue;
        if(k%n==0){
            ans++;
            while(k%n==0){
             k /= n;
            }
        }
        n += 2;
    }

    cout << ans << endl;
//  printf("%.10f\n",ans);

    return 0;
}