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 Go (1.6) (judge verdict: Runtime Error)

package main

import "fmt"

func gcd(x, y int) int {
	for y != 0 {
		x, y = y, x%y
	}
	return x
}

func main() {
	var a, b int
	fmt.Scan(&a, &b)

	g := gcd(a, b)
	ans := make([]int, 1, g)
	ans[0] = 1
	for i := 2; i <= g; i++ {
		isOk := true
		for _, v := range ans {
			if gcd(i, v) != 1 {
				isOk = false
				break
			}
		}
		if !isOk || gcd(i, a) == 1 || gcd(i, b) == 1 {
			continue
		}
		ans = append(ans, i)
	}
	fmt.Println(len(ans))
}