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: Time Limit Exceeded)

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

int main(void){

  int A = 0;
  int B = 0;
  scanf("%d\n",&A);
  scanf("%d\n",&B);
  int loopCount = 0;
  int resultCount = 0;

  if (A <= B) {
    loopCount = A / 2;
  }else{
    loopCount = B / 2;
  }
  if (A == 1 || B == 1) {
    printf("1\n");
  }else if(A == 2 || B == 2){
    printf("2\n");
  }else{
    int k;
  	for(long long int i = 3;i <= 10000000000;i += 2){
  		k = 0;
  		for(long long int j = 3;j <= sqrt(i);j += 2){
  			if(i % j == 0){
  				k = 1;
  				break;
  			}
  		}

  		if(k == 0){
        if (A % i == 0 && B % i  == 0) {
          resultCount++;
        }
      }
  	}
    printf("%d\n",resultCount + 2);
  }
}