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 <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <utility>
#include <math.h>
#include <stdexcept>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string.h>
#include <cassert>
#include <string.h>
#include <tuple>

using namespace std;
#define  rep(i,n) for(int i=0;i<n;i++)
#define  rrep(i,n) for(int i=n-1;i>=0;i--)///n-1番目からn回
#define  REP(i,a,b) for(int i=a;i<=b;i++) ///a番目からb回
#define  RREP(i,a,b) for(int i=a;i<b;i--)
typedef long long int ll;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pair_t;
#define MOD 1000000007

//int arr[N]
////sort(arr,arr +N)

//vector <int> hoge
////sort(hoge.begin(),hoge.end());
///*min_element(hoge.begin(),hoge.end();　最小値　　　最大値はmax_element();
///(10-(n%10))%10  10-(1桁目)取得 12->8 20->0

#define Ru(n)  ( ((n)+9)/10*10  ) //切りあげ
#define Rd(n)  ( ((n)-10)/10*10 +10 )	切り捨て


int gcd(int a,int b);
size_t length(const std::string& str) {
	return str.length();
}
bool comp(const pair_t &a, const pair_t &b) {
	return a.first < b.first;
}

int main() {
	long a,b,GCD,ans=0;
	cin>>a>>b;
	GCD=gcd(a,b);
	
	int prime[50]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,87,89,97};//26
	for(int i=0;i<26;i++){
		if(GCD%prime[i]==0)
			ans++;
	}
	for(long i=101;i<GCD;i+=2){
		if(GCD%i==0)
			ans++;

	}
	cout<<ans+1<<endl;

	return 0;

}


int gcd(int a,int b){
	if(b==0) return a;
	return gcd(b,a%b);
}