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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii  = pair<int,int>;
#define rep(i,n) for(int i=0; i<(n); i++)
#define Rep(i,n) for(int i=0; i<(n); i+=2)
#define rep2(i,x,n) for(int i=x; i<=(n); i++)
#define MOD 1000000007
#define EPS 10e-8
const int INF = 1e9;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; } //最大公約数
int lcm(int a, int b) {return a * b / gcd(a, b);}   //最小公倍数

bool compare_by_b(pair<int, ll> a, pair<int, ll> b)
{
    if(a.second != b.second){
        return a.second > b.second;
    }else{
        return a.first < b.first;
    }
}
bool IsPrime(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(){
int a,b,c,i;

scanf("%d %d",&a,&b);
vector<int> v;
int count = 0;
int ans = 0;
c = a;
if(a>b)c=b;
    for(i=2;i<=c;i++){
    if(a%i==0&&b%i==0){
        if(IsPrime(i))
            ans++;
    }
    }

    cout << ans+1 << endl;
    return 0;
}