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

int X[100], Y[100];


// 最大公約数を求める
long long int GCD (long long int a, long long int b) {
    long long int tmp, x, r;
    if (a<b) {
        tmp = a;
        a = b;
        b = tmp;
    }
    r = a % b;
    while (r!=0) {
      a = b;
      b = r;
      r = a % b;
    }
    return b;
}


int sib(long long A) {
    int i;
    int num;
    
    num=0;
    if (A%2==0) num++;
    if (A%3==0) num++;
    while (A%2==0) {A/=2;}
    while (A%3==0) {A/=3;}
    
    for (i=5;i<sqrt(A);i+=4) {
        if (A%i==0) num++;
        while (A%i==0) {A/=i;}
        i+=2;
        if (A%i==0) num++;
        while (A%i==0) {A/=i;}
    }
    if (A!=1) num++;
    return num;
}


int main() {
  long long A, B, G, result;
  int i;
  scanf ("%lld %lld", &A, &B);
  
  G = GCD(A,B);
//  printf("G=%lld\n",G);
  
  result = sib(G);
  
  printf("%lld", result);
  return 0;
}