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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <numeric>
using namespace std;
typedef long long int ll;

#define EPS (1e-7)
#define INF 1e18
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define PI (acos(-1))

#define REP(i, n) for(int i = 0; i < (int)(n); i++)

int gcd(int a, int b){
    if (a < b) return gcd(b, a);
    if (b == 0) return a;
    return gcd(b, a % b);
}

bool is_prime(int x){
    double tmp = sqrt(x) + 1;
    if (x == 1) return true;
    REP(i, tmp){
        if (i < 2) continue;
        if (x % i == 0) return false;
    }
    return true; 
}

int main() {

    ll A, B;
    cin >> A >> B;

    int g = gcd(A, B);
    double tmp = sqrt(g) + 1;
    int ans = 1;
    REP(i, tmp){
        if (i < 2) continue;
        if (i > g) break;
        if (g % i == 0){
            //cout << i << endl;
            ans++;
            while (g % i == 0) g /= i;
        }
    }
    if (g != 1) ans++;
    cout << ans << endl;


    return 0;
}