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: Wrong Answer)

#include <stdio.h>
int main(){
	long long int a, b;
	long long int c=1;
	long long int max, min;
	scanf("%lld %lld",&a,&b);
	max = b;
	min = a;
	if(a>b) {
		max = a;
		min = b;
	}
	long long int f = 2;
	long long int n = 2;
	
	while(1){
		if(max%f==0 && min%f==0){
			while(max%f==0 && min%f==0){
				max = max/f;
				min = min/f;
			}
			c++;
		}
		if(min%f < min) break;
		f = 2*n - 1;
		n++;
	}
	
	
	printf("%lld\n",c);
	return 0;
}