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

int main() {
	long int a, b,max,min,count=1,count2=0;
	long int history[10000] = { 1 }, history2[10000] = { 1 };
	int flag = 1;

	scanf("%ld %ld", &a, &b);
	if (a >= b)max = a,min=b;
	else max = b,min=a;

	for (int i=2;i<=sqrt(max);i++){
		if (count == 1)flag = 1;
		else {
			for (int j = 1; j < count; j++) {
				if (i % history[j] == 0) {
					flag = 0;
					break;
				}
			}
		}
		if ((a % i == 0) && (b % i == 0)&&(flag!=0)) {
			history[count] = i;
			history2[count] = max / i;
			printf("history2[%d]=%d\n", count, history2[count]);
			count++;
		}
		flag = 1;
	}
	printf("count=%d\n", count);
	for (int i = 1; i <= count-1; i++) {
		for (int j = 1; j <= count-1; j++) {
			if ((history2[i] % history[j] == 0)){
				flag = 0;
				break;
			}
		}
		if ((flag == 1) && (a % history2[i] == 0) && (b % history2[i] == 0))count2++;
		flag = 1;
	}
	if ((count == 1) && (max % min==0)&&(min!=1)) {
		count++;
	}
	printf("%d", count+count2);
	return 0;
}