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>

int main(void){
	int A,B;
	int cd[10000];
	scanf("%d %d",&A,&B);
	int i,j;
	int flag,c;
	int min = (A>B) ? B:A;
	for(i=0; i<min; i++) cd[i]=0;
	c=0;
	for(i=1; i<=min; i++){
		flag=0;
		if(A%i==0 && B%i==0) flag=1;
		if(flag==1){
			for(j=1; j<c; j++){ 
				if(i%cd[j]==0){flag=0; break;}
			 }
		}
		if(flag==1){ cd[c]=i; c++; } 
	}
	printf("%d\n", c);
}