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 <bits/stdc++.h>

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = ((int)(n)-1); i >= 0; i--)
#define all(x) (x).begin(),(x).end()
 
using namespace std;
using ll = long long;
 
struct edge { int to, cost; };
 
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> P;
 
const int INF = ~(1<<31);
const int MOD = 1e9+7;
const double EPS = 1.0e-10;
const double PI = acos(-1.0);
 
const int mx4[] = {0,1,0,-1};
const int my4[] = {1,0,-1,0};
const int mx8[] = {0,1,1,1,0,-1,-1,-1};
const int my8[] = {1,1,0,-1,-1,-1,0,1};

int main()
{
	ll a,b,ta,tb;
	cin >> a >> b;
	ta = a,tb = b;
	map<ll,ll> mp;
	mp[1]++;
	for(ll i = 2; i*i <= max(ta,tb); i++){
		while(a%i == 0 && b%i == 0){
			a /= i;
			b /= i;
			mp[i]++;
		}	
	}

	if(a < b) swap(a,b);
	if(a%b == 0) mp[a]++;

	cout << mp.size() << endl;
 	return 0;
}