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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <cmath>

using namespace std;

template <int base>
int digit_sum(int n)
{
	int sum = 0;
	while (n > 0) {
		sum += n % base;
		n /= base;
	}
	return sum;
}

using ivec = vector<int>;

ivec read_int_args()
{
	int N;
	cin >> N;
	ivec v(N);
	for (auto& e : v) {
		cin >> e;
	}
	return v;
}

int compar(const void * b, const void* a)
{
	return (*(int*)a - *(int*)b );
}

template<typename T> bool chmax(T& a, const T& b)
{
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

template<typename T> bool chmin(T& a, const T& b)
{
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}

long long gcd(long long a,long long b)
{
	if (a%b==0)
	{
		return(b);
    }
	else
	{
		return(gcd(b,a%b));
    }
}

bool func(long long &A, long long d)
{
	
	if ((A % d)) {
		return false;
	}
	A /= d;
	while (!(A % d)) {
		A /= d;
	}
//	cout << "T" << d << endl;
	return true;
}

int main()
{
	long long A, B;
	cin >> A >> B;
	long long g = gcd(A, B);
	int n = 0;
	// 2
	if (func(g, 2)) n++;
	if (func(g, 3)) n++;
	if (func(g, 5)) n++;
	
	long long d = 7;
	float sqg = sqrt(g);
	while (g < sqg && g > 1) {
		if (func(g, d)) {
			n++;
			sqg = sqrt(g);
		}
		d += 2;
	}

	cout << (n + 1) << endl;

	return 0;
}