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<stdbool.h>

typedef long long int ll;

static inline ll gcd(ll a,ll b){
    ll max,min,r;
    max = a<b?b:a;
    min = a<b?a:b;
    while(min){
        r = max;
        r %= min;
        max = min;
        min = r;
    }
    return max;
}

bool prime_flag[1000001];//caution true -> 0 fasle -> 1

int main(void){
    ll a,b,gcdnum,count = 1;
    scanf("%lld %lld",&a,&b);
    gcdnum = gcd(a,b);
    for(int i = 2;i*2 < 1000001;i++)for(int j = 2;i * j < 1000001;j++)prime_flag[i*j] = 1;
    for(ll i = 2;i*i <= gcdnum;i++)if(!(prime_flag[i] || (gcdnum % i))){
        count++;
        while(!(gcdnum % i))gcdnum /= i;
    }
    if(gcdnum != 1)count++;
    printf("%lld",count);
    return 0;
}