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: Time Limit Exceeded)

#include <stdio.h>

int main(void){
	
	long long A;
	long long B;
	long long C;
	long long D[10000];
	D[0] = 0;

	scanf("%lld", &A);
	scanf("%lld", &B);

	if(A < B){
		C = A;
	}else{
		C = B;
	}
	
	for(int i = C; i > 0; i--){
		if(A % i == 0 && B % i == 0){
			C = i;
			i = 0;
		}
	}

	int cnt = 1;
	D[cnt] = 1;
	for(int i = 2; i <= C; i++){
		if(A % i == 0 && B % i == 0){
			cnt++;
			D[cnt] = i;
		}
	}

	for(int i = 2; i <= cnt; i++){
		if(D[i] != 0){
			for(int j = i + 1; j <= cnt; j++){
				if(D[j] % D[i] == 0){
					D[j] = 0;
				}
			}
		}
	}

	int count = 0;
	for(int i = 1; i <= cnt; i++){
		if(D[i] != 0){
			count++;
		}
	}

	printf("%d\n", count);

	return 0;
}