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: Time Limit Exceeded)

#define _USE_MAT_DEFINES
#include<iostream>
#include <iomanip>
#include<stdio.h>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include <queue>
#include <utility>
#include <map>
#include <set>
#include <stack>
#include <deque>
#include <numeric>
#include <functional>
using namespace std;

static const int MOD = 1e9 + 7;

int GCD(int A, int B) {
    if (A == 1)return 1;
    if (A == 0)return B;
    else return GCD(B % A, A);
}

int NumberOfPrimeFactor(int x) {
    int cnt = 0;
    for (int i = 2; i*i <x ; i++){
        if (x%i == 0) {
            cnt++;
            while (x%i == 0)x /= i;
        }
    }
    return cnt + (x != 1);
}

int main() {

    int A,B;
    cin >> A >> B;

    if (A > B)swap(A, B);
    cout << NumberOfPrimeFactor(GCD(A, B)) + 1;

}