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>
static int a[100010],hantei[10010];

int gcd(int a, int b) {
  int t;    // �⏕�ϐ�
  while ( b != 0 ) {          // ������0�ɂȂ�܂�
    t = a%b; a = b; b = t;    // �������]�ɒu��������
  }
  return a;                   // ������0�ɂȂ����Ƃ��̈����GCD
}

int 
main(){
	 int A,B,i;
	 int cnt=0;
	int K;
	scanf("%d %d",&A,&B);

	int tmp;
	if(B>A){tmp = A;
			A=B;
			B=tmp;
	}
	
	
	if(B==1){
		printf("1\n");return 0;
	}
	
	for(i=1;i<B;i++){
		if(A%i==0 && B%i==0)a[cnt++]=i;
		
	}
	int bb,j;
	for(i=1;i<cnt;i++){
		//printf("%d : ",a[i]);
		for(j=0;j<i;j++){
			
			/*if(a[i]==2){
				hantei[a[i]]=1;
			}
			else if (a[i]>2 && a[i]%2==1){*/
				bb = gcd(a[j],a[i]);
						hantei[bb]=1;//}
			//printf("%d(%d,%d) ,,,",bb,a[j],a[i]);

		}
	}
	//printf("\n");
	int ans =0;
	//printf("%dlllll\n",a[1]);
	//if(a[1]==0)ans++;
	
	for(i=0;i<cnt;i++){
		if(hantei[a[i]]){
			 int k,flag=0;
			 for( k=2;k<a[i];++k) {
    			if( a[i]%k==0 ) {
      			flag = 1;
      			break;
    		}
		  }
			if(flag==0)ans++;

		}
		//printf("(%d %d:%d),", a[i],hantei[a[i]],ans);
	}
	printf("%d\n",ans);
 return 0;
}