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 C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <math.h>
#include <map>

long long euclid(long long a, long long b){
  long long red = a % b;
  if (red == 0){
    return b;
  } else {
    a = std::max(red, b);
    b = std::min(red, b);
    return euclid(a,b);
  }
}

bool is_prime(long long a){
    bool ans = true;
  for (long long i=2; i<=sqrt(a); i++){
    if (a%i==0){
      ans = false;
    }
  }
  return ans;
}

int main(){
  long long A, B;
  scanf("%lld %lld",&A, &B);
  long long c = euclid(std::max(A,B),std::min(A,B));
  if (is_prime(c)){
    printf("2\n");
    return 0;
  }
  //printf("c=%d\n",c);
  long long ans=1; //1は回答の前提としておく
  for (long long i=2; i<=c; i++){
    if (c%i==0){
      ans++;
      c = c/i;
      while(c%i==0){
        c = c/i;
      }
    }
  }
  printf("%lld\n",ans);
  return 0;
}