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 Python (3.4.3) (judge verdict: Accepted)

def find_lcd(A,B):
  if A==1 or B==1:
    return 1
  if A==B:
    return A
  else:
    return find_lcd(min(A,B), abs(A-B))
  
def find_num_prime(X):
  if X==1:
    return 1
  count = 1 # 1
  i=2
  sqrt_org=int(X**0.5)
  while X != 1:
    if i**2 > X:
      count += 1
      break
    if X%i == 0:
      count += 1
      while X%i == 0:
        #print(str(X)+' '+str(i))
        X = int(X/i)
    if i== 2:
      i += 1
    else:
      i += 2
  return count
  
A,B = map(int, input().split())
lcd=find_lcd(A,B)
#print(lcd)
print(find_num_prime(lcd))