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: Accepted)

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

//abc142D
int main(void) {
  int c=1;
  long long a, b, tmp, i=2, memo=1;
  scanf("%lld %lld", &a, &b);

  if(a < b) tmp=a, a=b, b=tmp; //ユークリッドの互除法//gcd(a,b)=>b
  while(a % b != 0){
    a=a % b;
    if(a < b) tmp=a, a=b, b=tmp;
  }
  //printf("b=%lld\n", b);
  while(i <= sqrt(b)){
    if(b % i == 0){
      if(memo != i) c++;
      b/=i, memo=i;
    }else i++;
  }
  if(b != 1 && b!=i) c++;
  printf("%d", c);
}