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<iostream>
#include<vector>
#include<cstdio>
#include<string>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;

using ll = long long;

int main()
{
    ll a, b;
    cin >> a >> b;

    ll large = max(a, b);
    ll small = min(a, b);

    a = small;
    b = large;

    vector<ll> al;
    vector<ll> bl;

    for (int i = 1; i <= sqrt(a); i++) {
        if (a % i == 0) {
            al.push_back(i);   
            if (i != sqrt(a)) {
                al.push_back(a / i);
            }
        }

    }
    for (int i = 1; i <= sqrt(b); i++) {
        if (b % i == 0) {
            bl.push_back(i);   
            if (i != sqrt(b)) {
                bl.push_back(b / i);
            }
        }
    }

    sort(al.begin(), al.end());
    sort(bl.begin(), bl.end());

    vector<ll> res;
    ll count = 0;
    for (int i = 0; i < al.size(); i++) {
        ll now = al[i];
        while(bl[count] < now) {
            count++;
            if (count >= bl.size()) {
                break;
            }
        }
        if (count >= bl.size()) {
            break;
        }
        if (bl[count] == now) {
            res.push_back(now);
        }
    }


    ll result = 0;
    for (int i = 0; i < res.size(); i++) {
        // 素数
        bool isSosu = true;
        for (int j = 2; j <= sqrt(res[i]); j++) {
            if (res[i] % j == 0) {
                isSosu = false;
                break;
            }
        }
        if (isSosu) result++;
    }

    cout << result << endl;
}