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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INF 10000000000
#define mod 1000000007
typedef long long int ll;
ll u(const void *a, const void *b){
    return *(ll*)a-*(ll*)b;
}
ll d(const void *a, const void *b){
    return *(ll*)b-*(ll*)a;
}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll gcd(ll a,ll b){
  if(!b)return a;
  return gcd(b,a%b);
}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
int f(ll a){
  for(int i=2;i<=sqrt(a);i++)if(a%i<1)return 0;
  return 1;
}
int main(void){
  ll a,b;
  scanf("%lld %lld",&a,&b);
  ll g=gcd(max(a,b),min(a,b)),ans=0;
  for(ll i=1;i<=sqrt(g);i++){
    if(g%i<1){
      if(f(i))ans++;
      if(i*i-g)if(f(g/i))ans++;
    }
  }
  PRL(ans);
}