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;
typedef long long ll;
using pint = pair<int,int>;
ll mod = 1000000007;
int dx[4] = {1,0,-1,0},dy[4] = {0,1,0,-1};
ll dp[100100][13];

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

int main() {
    ll x,y;
    cin >> x >> y;
    ll g = gcd(x,y);
    ll cnt = 1;
    for (ll i = 2; i*i < g; ++i) {
        if(g%i == 0){
            cnt++;
            while(g%i == 0){
                g /= i;
            }
        }
    }
    if(g != 1) cnt++;
    cout << cnt << endl;
    return 0;
}