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: Runtime Error)

from collections import defaultdict,deque, Counter
import sys,heapq,bisect,math,itertools,string,queue,copy,time
import numpy as np
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())
ans = 0

def num_check(n,a,b):
    just_list = list(range(2,n+1))
    ans_dq = deque([1])
    while len(just_list)!=0:
        first_val = just_list[0]
        ans_dq.append(first_val)
        just_list = [i for i in just_list if (i%first_val != 0 and a%i == 0 and b%i == 0)]
    return len(ans_dq)

def gcd(a,b):
    if b>a: return gcd(b,a)
    else: return gcd(b,a%b) if a%b else b

[a,b] = inpl()
n = gcd(a,b)
ans = num_check(n,a,b)
print(ans)