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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

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

struct ya{
		long long p;
		int co;
};

int main(){
		long long n,m,a;
		scanf("%lld%lld",&n,&m);
		a=gcd(n,m);
		int c=0,i;
		struct ya y[100000];
		y[0].co=0;
		for(i=2;i*i<=a;i+=2){
				if(a%i==0){
						y[c].p=i;
						while(a%i==0){
								a/=i;
								y[c].co++;
						}
	//					printf("%lld\n",a);
						c++;
						y[c].co=0;
				}
				if(i==2)i--;
		}if(a!=1)c++;
		printf("%d\n",c+1);
		return 0;
}