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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <set>
#include <algorithm>
 
using namespace std;
using uint = unsigned int;

uint gcd(uint a, uint b) {
    if(a < b) return gcd(b, a);
    else if(b == 0) return a;
    else return gcd(b, a%b);
}
 
set<uint> pf(const uint a) {
    set<uint> ans;
    uint t = a;
    
    for(uint i=2; i<=a; i++) {
        while(t % i == 0) {
            t /= i;
            ans.insert(i);
        }
    }
    
    return ans;
}
 
int main(void){
    uint A, B;
    cin >> A >> B;
    
    set<uint> abp = pf(gcd(A, B));
    
    cout << abp.size() + 1;
}