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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int main(void){

	long long int a,b;
	long long int num;
	long long int buff;
	int count_a;
	int count_b;
	long long int ret;

	
	int j;
	int i = 0;
	ret =0;
	
	scanf("%lld %lld",&a,&b);

	num = 2;

	if ( a > b){
		buff = a;
		a = b;
		b = buff;
	}

	count_a = 0;		count_b = 0;
	while(a % num == 0){
		count_a++;
		a /= num;
	}
//	while(b % num == 0){
//		count_b++;
//		b /= num;
//	}
	if(count_a > 0){
		ret++;
	}

//	if(count_a > 0 && count_b > 0){
//		ret++;
//	}

	for(num = 3;num <= a/2 && num <= b/2 ; num+=2){
		count_a = 0;		count_b = 0;
		while(a % num == 0){
			count_a++;
			a /= num;
		}
//		while(b % num == 0){
//			count_b++;
//			b /= num;
//		}

		if(count_a > 0){
			ret++;
		}
//		if(count_a > 0 && count_b > 0){
//			ret++;
//		}
	}
	if(a != 1 && b != 1){
		if(b % a == 0){
			ret++;
		}
	}


	printf("%lld\n",ret+1);



	return 0;
}