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: Runtime Error)

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

int main(){
  int aint, bint;
  scanf("%d %d", &aint, &bint);
  long long int a, b;
  a = aint;
  b = bint;
  if(a < b){
    int sw = a;
    a = b;
    b = sw;
  }
  int r = a % b;
  while(r != 0){
    a = b;
    b = r;
    r = a % b;
  }
  //gcd = b
  //printf("b = %d\n", b);
  //int rb = sqrt(b);
  int rb = b;
  int ans = 1;
  /*
  int e[rb];
  for(int i = 0; i < rb; i++){
    e[i] = 0;
  }
  */
  if(b % 2 == 0){
    ans++;
  }
  while(b % 2 == 0){
    b /= 2;
    //e[2] = 1;
  }
  for(int i = 3; i < rb; i += 2){
    if(b % i == 0){
      ans++;
    }
    while(b % i == 0){
      b /= i;
      //e[i] = 1;
    }
    if(b == 1){
      break;
    }
  }
  /*
  for(int i = 0; i < rb; i++){
    ans += e[i];
    //printf("%d ", e[i]);
  }
  */
  printf("%d\n", ans);
  return 0;
}