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

#define max(a,b) ((a)>(b))?(a):(b)
#define min(a,b) ((a)>(b))?(b):(a)

typedef long long ll;
typedef unsigned long long ull;

ll gcd(ll a, ll b)
{
	if (a % b) { return gcd(b, a % b); }
	return b;
}

int main(void)
{
	ll a, b;
	scanf("%lld%lld", &a, &b);
	a = gcd(a, b);

	ll p[64] = { 0 }, n = 2;
	int pnc[64] = { 0 };
	int pcnt = 0, cnt = 0, pnclen = 0;
	ll sqrt_a = sqrt(a);
	while (a != 1)
	{
		if (!(a % n))
		{
			a /= n;
			if (pcnt && p[pcnt - 1] != n)
			{
				pnclen++;
			}
			pnc[pnclen]++;
			p[pcnt++] = n;
		}
		else
		{
			n++;
			if (n>sqrt_a)
			{
				printf("1"); return 0;
			}
		}
	}
	if(pcnt) pnclen++;


	printf("%d", pnclen + 1);
}