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>

#define fti(i, x, y) for(int i = (x), _y = (y); i <= _y; ++ i)
#define ftd(i, x, y) for(int i = (x), _y = (y); i >= _y; -- i)
#define ll long long

using namespace std;

const int N = 1e6 + 2;
ll a, b;
int res, prime[N], cnt;
bool dd[N];

void inp()
{
    scanf("%lld %lld", &a, &b);
}

void snt()
{
    int j;
    for(int i = 2; i * i <= N - 2; ++ i) {
        j = 2;
        if(!dd[i]) {
            while(i * j <= N - 2) dd[i * j] = true, ++ j;
            prime[++ cnt] = i;
        }
    }
}

void proc()
{
    snt();
    ll tg = sqrt(min(a, b)), Min = min(a, b);
    fti(i, 1, tg) {
        if(a % i == 0 && b % i == 0) {
            if(!dd[i]) ++ res;
            if(Min / i == 1 || Min / i == i) goto x;
            fti(j, 1, cnt) {
                if(prime[j] > sqrt(Min / i)) break;
                if((Min / i) % prime[j] == 0) goto x;
            }
            ++ res;
            x:;
        }
    }
    printf("%d\n", res);
}

int main()
{
    inp();
    proc();
    return 0;
}