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 Python (3.4.3) (judge verdict: Wrong Answer)

from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush
from fractions import gcd
from copy import deepcopy
import math
import queue
#import numpy as np
#import sympy as syp(素因数分解とか)
Mod = 1000000007
 
  
def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))+1):
        if prime[i] == 1:
            for j in range(2*i,n+1):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n+1):
        if prime[i] == 1:
            res.append(i)
    return res

def factorial(i):
    if i == 1:
        return 1
    else:
        return i*factorial(i-1)
 
class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[py] = px
        else:
            self.parent[px] = py
 
    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)




def main(): #startline-------------------------------------------
    a,b = map(int,input().split())
    d = gcd(a,b)
    p = int(math.sqrt(d))+1
    div = []
    for i in range(1,p):
        if d%i==0:
            if d//i != i:
                div.append(i)
                div.append(d//i)
            else:
                div.append(i)
    div.sort()
    n = len(div)
    sgn = [1]*(n)
    for i in range(1,int(math.sqrt(n))):
        if sgn[i] == 1:
            for j in range(i+1,n):
                if div[j]%div[i] == 0:
                    sgn[j] = 0
    res = []
    for i in range(n):
        if sgn[i] == 1:
            res.append(div[i])
    print(len(res))
if __name__ == "__main__":
    main() #endline===============================================