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: Accepted)

fun main(args: Array<String>) {
    val (A, B) = listOfLong()
    val G = gcd(A, B)
    val C = factorize(G)
    println(C.count() + 1)
}
fun next() = readLine()!!
fun listOfString() = next().split(" ")
fun listOfLong(delta: Long = 0L) = listOfString().map { java.lang.Long.parseLong(it) + delta }
tailrec fun gcd(a: Long, b: Long): Long = if (a % b == 0L) b else gcd(b, a % b)
fun factorize(N: Long): MutableList<Long> {
    val res = mutableListOf<Long>()
    var n = N
    var i = 2L
    while (i * i <= n) {
        if (n % i == 0L) {
            var m = 0
            while (n % i == 0L) {
                n /= i
                m += 1
            }
            if (m > 0) res.add(i)
        }
        i += 1L
    }
    if (n != 1L) res.add(n)
    return res
}