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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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


int main(void)
{
	long int i, j;
	long int a, b;
	long int sum;
	long int max;
	long int count;
	long int flg;
	long int r;
	long int work;
	
	scanf("%ld%ld", &a, &b);
	
	if(a < b) {
		work = a;
		a = b;
		b = work;
	}
	
	max = a;
	
	r = a % b;
	while(r != 0) {
		a = b;
		b = r;
		r = a % b;
	}
	
	max = b;
	
	//bが最大公約数
	/*
	printf("%ld\n", b);
	
	return 0;
	*/
	
	flg = 1;
	
	if(b % 2 == 0) {
		count = 2;
		while(b % 2 == 0) {
			b /= 2;
		}
	} else {
		count = 1;
	}
	/*
	printf("%ld\n", b);
	
	return 0;
	*/
	while(flg == 1) {
		for(i = 3; i <= (int)sqrt(max); i = i + 2) {
			if(b == 1) {
				break;
			}
			if(b % i == 0) {
				count++;
				while(b % i == 0) {
					b /= i;
				}
			}
		}
		flg = 0;
	}
	
	printf("%ld\n", count);
	
	return 0;
}