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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main(){
  int a = 0;
  int b = 0;
  int prime[1000000] = {};
  int primecnt = 1;
  int primeflag = 0;
  int smaller = 0;
  int hoge = 1;
  int i = 0;
  int j = 0;
  prime[0] = 2;
  scanf("%d %d",&a,&b);
  if(a < b){
    smaller = a;
  }else{
    smaller = b;
  }
  
  for(i = 2;i <= smaller;i++){
    primeflag = 1;
    for(j = 0;j < primecnt;j++){
      if(i % prime[j] == 0){
        primeflag = 0;
        break;
      }
    }
    if(primeflag){
      prime[primecnt] = i;
      primecnt++;
    }
  }
  
  for(i = 0;i < primecnt;i++){
    if(a % prime[i] == 0 && b % prime[i] == 0){
      hoge++;
    }
  }
  printf("%d",hoge);
  return 0;
}