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<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<functional>
#include<utility>
using namespace std;
int main(){
	int a, b; cin >> a >> b;
	bool prime[1000001] = {};
	int x = min(a,b),cnt=0;
	vector<int>data;
	for (int i = 2; i <= x; ++i) prime[i] = true;
	for (int i = 2; i*i <= x; ++i)
		if (prime[i]) {
			for (int j = 2; i*j <= x; ++j) prime[i*j] = false;
		}
	for (int i = 2; i*i <= x; ++i) {
		if (a%i == 0) {
			int y = a / i;
			if (b%i == 0) data.push_back(i);
			if (b%y == 0) data.push_back(y);
		}
	}
	for (unsigned int i = 0; i < data.size(); ++i)
		if (prime[data[i]]) {
			++cnt;
		}
	cout << cnt+1 << endl;
	return 0;
}