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)

#include<bits/stdc++.h>
using namespace std;
long long a,b,c,d,e,p[1000009],pi,sac[1000009],i,j,pas;
int main(){
    ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>a>>b;
    for(i=2; i<=1000000; i++){
        if(sac[i]==0){
            pi++;
            p[pi]=i;
            for(j=i*i; j<=1000000; j+=i) sac[j]=1;
        }
    }
    c=a;d=b;
    for(i=1; i<=pi; i++){
        while(c%p[i]==0) c/=p[i];
        while(d%p[i]==0) d/=p[i];
        if(a%p[i]==0&&b%p[i]==0) pas++;
    }
    if(c!=1&&d!=1&&c==d) pas++;
    pas++;
    cout<<pas;
    return 0;
}