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;
typedef long long ll;
const int MAX=10000050;
int f[MAX],zs[MAX];
int a[MAX],b[MAX],t=1,cnt=0;
void qzs(void)
{
    f[1]=f[2]=0;
    zs[0]=1;
    for(int i=2;i<MAX;i++){
        for(int j=2;!f[i]&&j*i<MAX;j++)
            f[j*i]=1;
        if(!f[i])
            zs[t++]=i;
    }
}
long long fenjiea(int x)
{
    int i=1;
    while(x&&i<t)
    {
        while(x%zs[i]==0){
            x/=zs[i];
            a[i]++;
        }
        i++;
    }
    return x;
}
ll fenjieb(int x)
{
    int i=1,flag=0;;
    while(x&&i<t)
    {
        while(x%zs[i]==0){
            x/=zs[i];
            flag=1;
        }
        if(flag&&a[i])
            cnt++;
        flag=0;
        i++;
    }
    return x;
}
int main()
{
    qzs();
    //cout << zs[t-1] << endl;
    ll A,B;
    cin >> A >> B;
    ll aa=fenjiea(A),bb=fenjieb(B);
    if(aa==bb&&aa)
        cnt++;
    cout << cnt << endl;
    return 0;
}