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 int long long
const int maxm=1e5+5;



signed main(){
    int a,b;cin>>a>>b;
    map<int,int>mark;
    for(int i=2;i*i<=a;i++){
        if(a%i==0){
            mark[i]=1;
            while(a%i==0)a/=i;
        }
    }
    if(a!=1)mark[a]=1;
    //
    int ans=0;
    for(int i=2;i*i<=b;i++){
        if(b%i==0){
            if(mark[i])ans++;
            while(b%i==0)b/=i;
        }
    }
    if(b!=1){
        if(mark[b])ans++;
    }
    //
    ans++;
    cout<<ans<<endl;
    return 0;
}
/*


*/