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: Runtime Error)

function getPrimeFactors(n) {
    if (n < 2) return [];
    if (n === 2 || n === 3 || n === 5) return [n];
    if (n % 2 === 0) return [2, ...getPrimeFactors(n/2)];
    if (n % 3 === 0) return [3, ...getPrimeFactors(n/3)];
    if (n % 5 === 0) return [5, ...getPrimeFactors(n/5)];
    let prime = 7;
    let step = 4;
    const limit = Math.sqrt(n);
    while (prime <= limit) {
      if (n % prime === 0) return [prime, ...getPrimeFactors(n/prime)];
      prime += step;
      step = 6 - step;
    }
    return [n];
};

function main(input) {
    const args = input.split(' ')
    const n1 = args[0]
    const n2 = args[1]
    const a = getPrimeFactors(n1).filter(function (x, i, self) { return self.indexOf(x) === i;})
    const b = getPrimeFactors(n2).filter(function (x, i, self) { return self.indexOf(x) === i;})
    console.log(a.map( x => b.find( y => x == y)).filter( x => x != null).length + 1)
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));
// main(require('fs').readFileSync('./ans.txt', 'utf8'));