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: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

const pi = math.Pi

var mod int = pow(10, 9) + 7
var Umod uint64 = 1000000007
var ans, cnt, sum int = 0, 0, 0

func main() {
	reader.Split(bufio.ScanWords)
	A, _ := strconv.Atoi(read())
	B, _ := strconv.Atoi(read())

	cnt = gcd(A, B)
	for i := 2; (i-1)*(i-1) < gcd(A, B); i++ {
		if cnt%i == 0 {
			cnt /= i
			ans++
		}
		for cnt%i == 0 {
			cnt /= i
		}
	}
	fmt.Println(ans + 1)
}

/*  ----------------------------------------  */

var reader = bufio.NewScanner(os.Stdin)

func read() string {
	reader.Scan()
	return reader.Text()
}

func lcm(x, y int) int {
	return (x / gcd(x, y)) * y
}

func gcd(x, y int) int {
	if x%y == 0 {
		return y
	} else {
		r := x % y
		return gcd(y, r)
	}
}

var fac [1000000]int
var finv [1000000]int
var inv [1000000]int

func combination_init() {
	fac[0], fac[1] = 1, 1
	finv[0], finv[1] = 1, 1
	inv[1] = 1

	// invは a^(-1) mod p
	// pをaで割ることを考える

	// p/a*(a) + p%a = p
	// p/a*(a) + p%a = 0          (mod p)
	// -p%a = p/a*(a)             (mod p)
	// -p%a *a^(-1)= p/a          (mod p)
	// a^(-1)= p/a * (-p%a)^(-1)  (mod p)
	// a^(-1) =

	for i := 2; i < 1000000; i++ {
		fac[i] = fac[i-1] * i % mod
		inv[i] = mod - inv[mod%i]*(mod/i)%mod
		finv[i] = finv[i-1] * inv[i] % mod
	}
}

func combination(x, y int) int {
	if x < y {
		return 0
	}
	if fac[0] != 1 {
		combination_init()
	}
	return fac[x] * (finv[y] * finv[x-y] % mod) % mod
	//return fac[x] / (fac[y] * fac[x-y])
}

func permutation(x, y int) int {
	if x < y {
		return 0
	}
	if fac[0] != 1 {
		combination_init()
	}
	return fac[x] * (finv[x-y] % mod) % mod
	//return fac[x] / fac[x-y]
}

func max(x ...int) int {
	var res int = x[0]
	for i := 1; i < len(x); i++ {
		res = int(math.Max(float64(x[i]), float64(res)))
	}
	return res
}

func min(x ...int) int {
	var res int = x[0]
	for i := 1; i < len(x); i++ {
		res = int(math.Min(float64(x[i]), float64(res)))
	}
	return res
}
func pow(x, y int) int { return int(math.Pow(float64(x), float64(y))) }
func abs(x int) int    { return int(math.Abs(float64(x))) }
func floor(x int) int  { return int(math.Floor(float64(x))) }
func ceil(x int) int   { return int(math.Ceil(float64(x))) }

type SortBy []int

func (a SortBy) Len() int           { return len(a) }
func (a SortBy) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a SortBy) Less(i, j int) bool { return a[i] < a[j] }

type PriorityQueue []int

func (h PriorityQueue) Len() int            { return len(h) }
func (h PriorityQueue) Less(i, j int) bool  { return h[i] > h[j] }
func (h PriorityQueue) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *PriorityQueue) Push(x interface{}) { *h = append(*h, x.(int)) }
func (h *PriorityQueue) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}