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 <cmath>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;

inline void read(int &a)
{
    a=0;int c=getchar();
    while(c>'9'||c<'0') c=getchar();
    while(c>='0'&&c<='9') a=(a<<3)+(a<<1)+c-48,c=getchar();
}

const int N=1e8+5,M=1e5+5;
int n,m,prime[N],cnt,ans;
bool not_pr[N];
ll a,b;
inline void get_prime()
{
    not_pr[1]=1;
    for(int i=2;i<=n;i++)
    {
        if(!not_pr[i]) prime[++cnt]=i;
        for(int j=1;j<=cnt&&i*prime[j]<=n;j++)
        {
            not_pr[i*prime[j]]=1;
            if(i%prime[j]==0) break;
        }
    }
}

int main()
{
    n=100000000;
    get_prime();
    cin>>a>>b;
    prime[0]=1;
    for(int i=0;i<=cnt;i++)
    {
        if(a<prime[i]||b<prime[i]) break;
        if(a%prime[i]==0&&b%prime[i]==0) ans++;
    }
    cout<<ans;
}