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 <stdlib.h>
#include <math.h>
 

int getSosuu(long long int start,long long int end){
	long long int i,j,k;
	for(i=start;i<=end;i+=2){
      k=0;
      for(j=3;j<=sqrt(i);j+=2){
        if(i%j==0){
          k=1;
          break;
        }
      }
      if(k==0) return i;
	}
 
	return 0;
}

void findFact(long long int num,long long int result[]){
  long long int i, cnt=0;
  for (i = 2; i <= num; i++){
    i = getSosuu(i,1000000000000);
    while (num % i == 0){
    if(cnt==0||(cnt>0&&result[cnt-1]!=i)){
        result[cnt]=i;
        cnt+=1;
    }
    num /= i;
    }
  }
}
int main(void) {
 
  long long int a,b,cnt=0,cnt2=0,result=0,
                fact_a[10000],fact_b[10000];
  scanf("%lld %lld",&a,&b);

  findFact(a,fact_a);
  findFact(b,fact_b);

  if(a>=1&&b>=1)result=1;
  while(fact_a[cnt]!=0){
    if(fact_a[cnt]==fact_b[cnt2])result+=1;
    if(fact_a[cnt]<fact_b[cnt2]){
      cnt+=1;
    }else{
      cnt2+=1;
    }
    if(fact_b[cnt2]==0)break;
  }

  printf("%lld\n",result);

  return 0;
}