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)

// ABC-142 D:
// 2019.9.28 bal4u
// Template for Online Contest

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

typedef long long ll;

long long gcd(long long a, long long b)
{
	long long r;
	while (b != 0) r = a % b, a = b, b = r;
	return a;
}

#define MAX 20000
int sz = 0;
ll a[MAX];

void check(ll n) {
	int i;
	for (i = 0; i < sz; i++) {
		if (gcd(n, a[i]) > 1) return;
	}
	a[sz++] = n;
}

int main()
{
	ll A, B, g;
	int i, b;
	
	scanf("%lld%lld", &A, &B);
	g = gcd(A, B);
	b = (int)sqrt((double)g);
	for (i = 2; i <= b; i++) {
		if (g % i == 0) check((ll)i), check(g/i);
	}
	printf("%d\n", sz+1);
	return 0;
}