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: Time Limit Exceeded)

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

bool sosuJudge(unsigned long number){
    /* 判定フラグ */
    bool flag = false;
    unsigned long i;
    
    /* 素数かどうかを判定 */
    for( i=2;i <= (number >> 1);++i ) {
        if( number % i==0 ) {
            flag = true;
            break;
        }
    }
    
    if(flag == false){//素数
        return true;
    }else{
        return false;
    }
}

int main(void){
    unsigned long i,j;
    unsigned long A,B,index = 1;//,box[10000000]
    int answer = 0;
   
    scanf("%ld %ld",&A,&B);
    
    unsigned long a,b,c;
    a = A;
    b = B;
    c=a;
    if(a>b)c=b;
    for(i=2;i<=c;i++){
        if(a%i==0 && b%i==0 && sosuJudge(i)){
            //box[index] = i;
            index++;
            //printf("%lu\n",i);
        }
    }
    
    printf("%lu\n",index);
    
    return 0;
}