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 (1.3.71) (judge verdict: Wrong Answer)

fun main(args: Array<String>){
    var (A,B) = readLine()!!.split(" ").map{it.toLong()}
    var ans :Long = 1


    var i = 2L
    var aSet = HashSet<Long>()
    var tmpA = A
    while(i * i < A){
        if(tmpA % i == 0L){
            aSet.add(i)
            while(tmpA % i == 0L) tmpA /= i
        }
        i++
    }

    var bSet = HashSet<Long>()
    var tmpB = B
    var j = 2L
    while(j*j<B){
        if(tmpB % j == 0L){
            bSet.add(j)
            while(tmpB % j == 0L) tmpB /= j
        }
        j++
    }

    if(A > 1 && B >1 && (A % B == 0L || B % A == 0L)) {
        println(aSet.intersect(bSet).size + 2)
    } else {
        println(aSet.intersect(bSet).size + 1)
    }




}