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 Kotlin (1.0.0) (judge verdict: Runtime Error)

fun main(args: Array<String>){
  val input = readLine()!!.split(" ")
  val a = input[0].toInt()
  val b = input[1].toInt()

  val result = getCommonPrimeNumbers(a, b)

  println(result.keys.size+1)
}

fun primeFactorization(num: Int): Map<Int, Int>{
  val result = mutableMapOf<Int, Int>()

  var tmp = num
  var i = 2
  while( i <= tmp ){
    if( tmp / i * i == tmp ){
      if( result.containsKey(i) )
        result.put(i, result.get(i)!!+1)
      else
        result.put(i, 1)
      tmp /= i
      i = 1
    }
    i++
  }

  return result
}

fun getCommonPrimeNumbers(a: Int, b: Int): Map<Int, Int>{
  val primeNumbersA = primeFactorization(a)
  val primeNumbersB = primeFactorization(b)

  val result = mutableMapOf<Int, Int>()
  for( (k, v) in primeNumbersA )
    if( primeNumbersB.containsKey(k) )
      //result.put(k, if( v < primeNumbersB.get(k)!! ) v else primeNumbersB.get(k)!!)
      result.put(k, Math.min(v, primeNumbersB.get(k)!!))

  return result
}