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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

var rows = require("fs").readFileSync("/dev/stdin", "utf8").split("\n")

var A = parseFloat(rows[0].split(" ")[0])
var B = parseFloat(rows[0].split(" ")[1])

var isPrime = function (n) {
  if (n === 1) return false
  if (n === 2) return true
  if (n === 3) return true

  var sqrtn = Math.sqrt(n)

  for (var i = 2; i < sqrtn + 1; i++) {
    if (n % i === 0) return false
  }

  return true
}
var getPrimeFactors = function (n) {
  var sqrtn = Math.sqrt(n)
  var res = {}

  for (var i = 2; i < sqrtn + 1; i++) {
    if (n % i !== 0) continue

    var I = parseInt(n / i, 10)

    if (isPrime(i)) res[i] = true
    if (isPrime(I)) res[I] = true
  }

  return res
}
var getIntersection = function (S1, S2) {
  var res = {}
  var Base = Object.keys(S1) < Object.keys(S2) ? S1 : S2

  for (var key in Base) {
    if (S1[key] && S2[key]) res[key] = true
  }

  return res
}
var size = function (S) {
  return Object.keys(S).length
}
var commonPrimeFactors = getIntersection(getPrimeFactors(A), getPrimeFactors(B))

console.log(size(commonPrimeFactors) + 1)