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>

int sosu(int n){
  int i;

  /* 自然数 */

  /* 判定フラグ */
  int flag = 0;

  /* 自然数の入力 */

  /* 素数かどうかを判定 */
  for( i=2;i<n;++i ) {
    if( n%i==0 ) {
      flag = 1;
      break;
    }
  }

  /* 判定結果を出力 */
  if( flag==0 ){//素数
    return 1;
  }else{
    return -1;
 }
  return 0;
}



int main(){
  int a,b;
  scanf("%d %d",&a,&b);
  if(b<a){
    int tmp;
    tmp=a;
    a=b;
    b=tmp;
  }
  int sum=1;
  int i,k;
  for(i=2;i<=a;i++){
    k=sosu(i);
    if(k==1){
      if((b-a)%i==0){
        sum++;
      while(a%i==0){
        a=a/i;
      }
      }
    }
  }
  printf("%d",sum);
  return 0;
}