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: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<vector>
#include<functional>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<string.h>
#define _USE_MATH_DEFINES
#include<math.h>
#include<deque>
#include<set>
#include<map>
#include<queue>
#include<iostream>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(auto i=a;i<b;i++)
#define all(_x) _x.begin(), _x.end()
#define r_sort(_x) sort(_x.begin(), _x.end(), std::greater<int>())
#define vec_cnt(_a, _n) (upper_bound(all(_a), _n) - lower_bound(all(_a), _n))
ll gcd(ll a, ll b) { return a % b == 0 ? b : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
#define INF 1 << 30
const ll mod = 1000000007;
vector<vector<int>> G;
vector<int> K;
int main() {
	ll A, B, G, k, ans = 0;
	cin >> A >> B;
	G = gcd(A, B);
	k = (ll)sqrt(G);
	rep(i, 2, k + 1) {
		if (G % i == 0)ans++;
		while (G % i == 0)G /= i;
	}
	if (G > k)ans++;
	printf("%lld\n", ans + 1);
	return 0;
}