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)

fun main(args: Array<String>) {
    abc142d()
}

fun abc142d() {
    val (a, b) = readLine()!!.split(" ").map { it.toLong() }
    val gcd = gcd(a, b)
    val primes = mutableSetOf<Long>()
    isPrimeCommonDivider(2L, primes, a, b)
    for (l in 3L..Math.sqrt(gcd.toDouble()).toLong() + 1 step 2) {
        isPrimeCommonDivider(l, primes, a, b)
    }
    primes.add(1L)
    println(primes.count())
}

fun gcd(a: Long, b: Long): Long {
    //    var source = max(a, b)
    //    var div = min(a, b)
    var source = if (a > b) a else b
    var div = if (b <= a) b else a
    var r = source % div
    while (r != 0L) {
        source = div
        div = r
        r = source % div
    }
    return div
}

fun isPrimeCommonDivider(n: Long, primes: MutableSet<Long>, a: Long, b: Long): Boolean {
    if (a % n == 0L && b % n == 0L) {
        if (a in primes) return false
        for (p in primes) {
            if (n % p == 0L) return false
        }
        primes.add(n)
        return true
    } else {
        return false
    }
}