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 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <vector>
#include <cmath>

using namespace std;

bool search(vector<int> l, int n) {
    for (int i = 0; i < l.size(); i++) {
        if (l[i] != 1) {
            if (n % l[i] == 0) {
                return false;
            }
        }
    }
    return true;
}

bool search_if_exist(vector<int> l, int n) {
    for (int i = 0; i < l.size(); i++) {
        if (l[i] == n) {
            return true;
        }
    }
    return false;
}

#include <math.h> 

bool isPrime(long int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}

int main() {
    long int a, b;
    scanf("%ld%ld", &a, &b);
    int max_divisor;
    if (sqrt(a) > sqrt(b)) {
        max_divisor = sqrt(a);
    } else {
        max_divisor = sqrt(b);
    }

    vector<int> common_divisors(1, 1);
    bool is_ok, result;
    for (int i = 2; i < max_divisor; i++) {
        if (a % i == 0 && b % i == 0) {
            result = search_if_exist(common_divisors, i);
            if (!result) {
                is_ok = search(common_divisors, i);
                if (is_ok) {
                    common_divisors.resize(common_divisors.size() + 1);
                    common_divisors[common_divisors.size() - 1] = i;
                }
            }
            if (a % (a / i) == 0 && b % (a / i) == 0) {
                result = search_if_exist(common_divisors, (a / i));
                if (!result) {
                    is_ok = search(common_divisors, (a / i));
                    if (is_ok) {
                        common_divisors.resize(common_divisors.size() + 1);
                        common_divisors[common_divisors.size() - 1] = (a / i);
                    }
                }
            }
            if (a % (b / i) == 0 && b % (b / i) == 0) {
                result = search_if_exist(common_divisors, (b / i));
                if (!result) {
                    is_ok = search(common_divisors, (b / i));
                    if (is_ok) {
                        common_divisors.resize(common_divisors.size() + 1);
                        common_divisors[common_divisors.size() - 1] = (b / i);
                    }
                }
            }
        }
    }
    int num = common_divisors.size();
    if ((isPrime(a) && b % a == 0) || (isPrime(b) && a % b == 0)) {
        num++;
    }
    printf("%d", num);
    return 0;
}