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)

tailrec fun gcd(a: Long, b: Long): Long = if (b == 0L) a else gcd(b, a % b)

fun factorize(n0:Long):List<Pair<Long,Int>>{
    val res = mutableListOf<Pair<Long, Int>>()
    var i = 2L
    var n = n0
    while (i * i <= n) {
        if (n % i != 0L){
            i++
            continue
        }
        var c = 0
        while ( n % i == 0L){
            n /= i
            c++
        }
        res.add(i to c)
        i ++
    }
    if (n != 1L){
        res.add(n to 1)
    }
    return res
}


fun calc(a:Long, b:Long):Int = factorize(gcd(a, b)).toSet().size

fun main(args:Array<String>) {
    val (a, b) = readLine()!!.split(" ").map { it.toLong() }
    val ans = calc(a, b)
    println(ans)
}