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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def isDivisor?(a, b, i)
  a % i == 0 && b % i == 0
end

def getDivisors(a, b)
  d = []
  for i in 1..([a, b].min)
    if isDivisor?(a, b, i)
      d.push(i)
    end
  end
  return d
end

def isDisjoint?(a, b)
  cnt = 0
  for i in 1..([a, b].min)
    if isDivisor?(a, b, i)
      cnt += 1
    end
  end
  if cnt == 1
    return true
  else
    return false
  end
end

a,b = gets.chomp.split(' ').map(&:to_i)
d = getDivisors(a, b)
disjoint = []
count = 0


for i in 0..(d.length - 1)
  flag = 0
  for k in 0..(disjoint.length - 1)
    if !isDisjoint?(d[i], disjoint[k])
      flag = 1
    end
  end
  if flag == 0
    disjoint.push(d[i])
  end
end
p disjoint.length