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)

//* AuThOr GaRyMr *//
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define niv vector<int>
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define KEEP while(1)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define ff fflush(stdout)
using namespace std;
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
typedef pair<mp,mp> superpair;
bool Is(LL x){
	for(LL i=2;i*i<=x;i++){
		if(x%i==0) return 0;
	}
	return 1;
}
map<LL,int> cnt,cnt2;
int main(){
	LL a,b;
	cin>>a>>b;
	for(LL i=1;i*i<=a;i++) {
		if(a%i==0){
			if(Is(i)){
				cnt[i]=1;
			}
			if(Is(a/i)){
				cnt[a/i]=1;
			}
		}
	}
	for(LL i=1;i*i<=b;i++) {
		if(b%i==0){
			if(Is(i)){
				cnt2[i]=1;
			}
			if(Is(b/i)){
				cnt2[b/i]=1;
			}
		}
	}
	int res=0;
	for(map<LL,int> :: IT ite=cnt.begin();ite!=cnt.end();ite++){
		if(cnt2[ite->FIR]) res++;
	}
	cout<<res<<endl;
	return 0;
}