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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

A, B = map(int, input().split())

divisors = []
for i in range(1, min(A,B) + 1):
	if A % i == 0 and B % i == 0:
		divisors.append(i)


done = {}
def is_coprime(a,b):
	if (a,b) in done:
		return done[a,b]
	if b == 1:
		return True
	for i in range(2, min(A,B)):
		if a % i == 0 and b % i == 0:
			done[(a,b)] = False
			return False
	done[(a,b)] = True
	return True


def getMaxLen(sofar, divisors):
	if len(divisors) == 0:
		return len(sofar)
	curr_max = getMaxLen(sofar,divisors[1:])
	for i in range(len(divisors)):
		d = divisors[i]
		flag = False
		for s in sofar:
			if not is_coprime(d,s):
				flag = True
				break
		if flag:
			continue
		new = sofar.copy()
		
		new.append(d)
		curr_max = max(curr_max, getMaxLen(new, divisors[i+1:]))
	return curr_max

print(getMaxLen([], divisors))