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: Runtime Error)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int main(){
	long int a, b;
	scanf("%ld%ld", &a, &b);
	long int cd[50000];
	int ler;
	if(a>b) ler = b;
	else ler=a;
	int j=0;
	for(int i=1; i<=ler/2+1; i++){
		if((a%i==0) && (b%i==0)){
			cd[j] = i;
			j++;
		}
	}
	if(a>b){
		if(a%ler==0){
			cd[j]=ler;
		}
		else j--;
	}
	else{
		if(b%ler==0){
			cd[j]=ler;
		}
		else j--;
	}
	
	/*
	for(int i=0; i<=j; i++){
		printf("%d ", cd[i]);
	}
	*/
	
	int count=1;
	for(int i=1; i<=j; i++){
		if(cd[i]!=1){
			for(int k=i+1; k<=j; k++){
				if(cd[k]%cd[i]==0) cd[k]=1;
			}
			count++;
		}
	}
	
	printf("%d", count);
	
	
	return 0;
}