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)

#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<queue>
#define _USE_MATH_DEFINES
#include<math.h>
#include<cmath>
#include<ctime>
#include<map>
#define R(i, n) for(int i = 0;i < n;i++)
#define S(a) scanf("%d", &a)
#define S2(a, b) scanf("%d%d", &a, &b)
#define S3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define SL(a) scanf("%lld", &a)
#define SL2(a, b) scanf("%lld%lld", &a, &b)
#define SL3(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)
#define SA(i, n, a) R(i, n) scanf("%d", a + i)
#define SAL(i, n, a) R(i, n) scanf("%lld", a + i)
#define P(a) printf("%d", a)
#define PL(a) printf("%lld", a)
using namespace std;
using ll = long long;
using PR = pair<int, int>;
const ll MOD = 1e+9 + 7;
const int INF = 1e+9;

ll A, B;

ll gcd(ll x, ll y) {
	if (y == 0) return x;
	return gcd(y, x % y);
}

map< ll, int > primeFactor(ll n) {
    map< ll, int > ret;
    for (int i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            ret[i]++;
            n /= i;
        }
    }
    if (n != 1) ret[n] = 1;
    return ret;
}

int main() {
	SL2(A, B);
	ll g_num = gcd(A, B);
    map<ll, int> ans =primeFactor(g_num);
    P(ans.size() + 1);
}