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: Time Limit Exceeded)

package main

import (
	"fmt"
)

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

	// 最大公約数を求める
	r := a % b
	for r != 0 {
		a = b
		b = r
		r = a % b
	}

	if b == 1 {
		fmt.Println(1)
		return
	}
	tmp := b
	// 因数分解していく
	var divisors []int
	for i := 2; i <= b; i++ {
		if tmp%i == 0 {
			divisors = append(divisors, i)
		} else {
			continue
		}
		for tmp%i == 0 {
			tmp = tmp / i
		}
	}
	fmt.Println(len(divisors) + 1)
}