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: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val a = sc.nextLong()
    val b = sc.nextLong()
    println(problem142d(a, b))
}

fun problem142d(a: Long, b: Long): Int {
    /**
     * 素因数分解
     * 対象の値が1になるまで素数で順番に割っていき割った数を列挙する
     */
    fun computePrimeFactorList(a: Long, b: Long): List<Long> {
        val p = mutableListOf<Long>()
        var primeFactorA = a
        var primeFactorB = b
        var i = 2L
        while (primeFactorA % i == 0L && primeFactorB % i == 0L) {
            primeFactorA /= i
            primeFactorB /= i
            p.add(i)
        }
        i++
        while (i < Math.max(a, b)) {
            while (primeFactorA % i == 0L && primeFactorB % i == 0L) {
                primeFactorA /= i
                primeFactorB /= i
                p.add(i)
            }
            if (primeFactorA == 1L || primeFactorB == 1L) break
            i += 2
        }
        return p
    }
    return computePrimeFactorList(a, b).distinct().size + 1
}