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 (1.14.1) (judge verdict: Accepted)

package main

import "fmt"

func main() {
	var A, B int
	fmt.Scan(&A, &B)

	am := make(map[int]int)
	bm := make(map[int]int)

	a := A
	for i := 2; i*i <= A; i++ {
		if a%i != 0 {
			continue
		}
		for a%i == 0 {
			am[i]++
			a /= i
		}
	}
	if a != 1 {
		am[a]++
	}
	b := B
	for i := 2; i*i <= B; i++ {
		if b%i != 0 {
			continue
		}
		for b%i == 0 {
			bm[i]++
			b /= i
		}
	}
	if b != 1 {
		bm[b]++
	}
	count := 1
	for k := range am {
		if _, ok := bm[k]; ok {
			count++
		}
	}
	fmt.Println(count)
}