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 Lua (5.3.2) (judge verdict: Time Limit Exceeded)

local x,y = io.read("*l"):match("(%d+).(%d+)")
local numX,numY = tonumber(x),tonumber(y)
local result = 1

local function FindFactors(numZ)
  local arr = {}
  for i = 2, numZ do
    local check = false
    while numZ % i == 0 do
      numZ = numZ // i
      check = true
    end
    if check then
      table.insert(arr,1,i)
    end
  end
  return arr
end

local arrX = FindFactors(numX)
local arrY = FindFactors(numY)

for i=1,#arrX do
  for j=1,#arrY do
    if arrX[i] == arrY[j] then
      result = result + 1
    end
  end
end
print(result)