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)

#include "bits/stdc++.h"
using namespace std;
using ll=long long;
#define print(n) cout<<n<<endl
#define rep(i,a,n) for(int i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
#define P pair<ll,ll>
#define vl vector<ll>
#define vi vector<int>
#define vvl vector<vector<ll>>
#define vvi vector<vector<int>>
const int M=100111;
const int inf=1000000007;
const long long INF=1e18;
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
int ddx[8]={1,-1,0,0,1,-1,-1,1},ddy[8]={0,0,1,-1,1,-1,1,-1};


/*
vvl comb(int n, int r) {
	vvl v(n + 1,vl(n + 1, 0));
	for (int i = 0; i < v.size(); i++) {
		v[i][0] = 1;
		v[i][i] = 1;
	}
	for (int j = 1; j < v.size(); j++) {
		for (int k = 1; k < j; k++) {
			v[j][k] = (v[j - 1][k - 1] + v[j - 1][k]);
		}
	}
	return v;
}
*/

void sosu(ll x,bool f[]){
	ll y=x;
	for(ll i=2;i<=sqrt(y);i++){
		if(x%i==0){
			x/=i;
			f[i]=true;
			i--;
		}
	}
}

ll __gcd(ll x,ll y){
	if(x<y)swap(x,y);
	while(1){
		ll tmp=x%y;
		x=y;
		y=tmp;
		if(y==0)break;
	}
	return x;
}

	

int main(){
	ll a,b;
	cin>>a>>b;
	ll gcd=__gcd(a,b);
	int cnt=0;
	ll G=gcd;
	for(ll i=2; i*i<=G;i++){
		if(gcd%i==0)cnt++;
		while(gcd%i==0){
			gcd/=i;
		}
	}
	if(gcd!=1) cnt++;
	cout<<cnt+1<<endl;
}