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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.io.StdIn._

object Main extends App {
  val Array(a, b) = readLine.split(" ").map(_.toLong)

  def gcd(l: Long, r: Long): Long = if (r == 0) l else gcd(r, l % r)

  def div(n: Long, d: Int, cnt: Int): Int = {
    if (n < d * d) {
      cnt + (if (n == 1) 1 else 2)
    } else {
      def divP(l: Long): Long = if (l % d == 0) divP(l / d) else l
      divP(n) match {
        case `n` => div(n, d + 1, cnt)
        case other => div(other, d + 1, cnt + 1)
      }
    }
  }

  println(div(gcd(a, b), 2, 0))
}