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: Runtime Error)

object Main {
  def main(args:Array[String]) = {
    val sc = new java.util.Scanner(System.in)
    val a, b = sc.nextInt
    // 最大公約数
    val g = gcd(a, b)
    // 最大公約数までの素数群
    val ps = primes(g)
    // 素因数分解
    var s = Set[Int]()
    var current_g = g
    for(p<-ps if current_g != 1) {
      while(current_g % p == 0) {
        // 割り切れるだけ割る
        current_g = current_g / p
        s += p // 割った数を追加
      }
    }
    println(s.size+1)
    
  }
  def gcd(a:Int, b:Int): Int = {
    var x = Math.max(a,b)
    var y = Math.min(a,b)
    var m = x % y
    while(m!=0) {
      x = y
      y = m
      m = x % y
    }
    return y
  }
  def primes(max:Int): Array[Int] = {
    if(max<2) {
      return Array[Int]()
    }
    val ar = new Array[Int](max)
    for(i <- 2 to max-1) {
      for(j <- i*2 to (max-1, i) if ar(j) == 0) {
        ar(j) = 1
      }
    }
    ar.zipWithIndex.filter{case(v,_)=>v==0}.map{case(_,idx)=>idx}.tail.tail // 0と1を除外
  }
}