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: Wrong Answer)

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(n: Long): List<Long> {
        val p = mutableListOf<Long>()
        var primeFactor = n
        val prev = mutableListOf(primeFactor)
        for (i in 2..n) {
            while (primeFactor % i == 0L) {
                primeFactor /= i
                p.add(i)
            }
            if (i > 10000 && prev.takeLast(1000).distinct().size == 1) break
            prev.add(primeFactor)
        }
        return p
    }

    val aFactor = computePrimeFactorList(a)
    val bFactor = computePrimeFactorList(b)

    val aMap = hashMapOf<Long, Int>()
    aFactor.forEach {
        aMap[it] = aMap[it]?.plus(1) ?: 1
    }
    val bMap = hashMapOf<Long, Int>()
    bFactor.forEach {
        bMap[it] = bMap[it]?.plus(1) ?: 1
    }
    val list = mutableListOf<Long>(1)
    aMap.forEach {
        if (bMap.containsKey(it.key)) list.add(it.key)
    }
    return list.size
}