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>
#include<stdbool.h>

#define MAX(a,b) ((a)<(b)?(b):(a))
#define MIN(a,b) ((a)<(b)?(a):(b))

bool    prime_flag[1000000];

static inline long long gcd(long long a,long long b){
    long long max,min,r;
    max = MAX(a,b);
    min = MIN(a,b);
    while(min != 0){
        r = max;
        r %= min;
        if(min>r){max=min;min=r;}
        else max = r;
    }
    return max;
}

int main(void){
    long long a,b,gcdnum;
    scanf("%lld %lld",&a,&b);
    gcdnum = gcd(a,b);
    long long ans = 1;
    prime_flag[0] = false;
    for(long i = 1;i < 1000000;i++)prime_flag[i] = true;
    for(long  i = 1;i < 1000000;i++)if(prime_flag[i])
        for(long j = 2;(i+1)*j-1 < 1000000;j++)prime_flag[(i+1)*j-1] = false;
    for(long i = 0;i*i <= MIN(gcdnum,1000000000000);i++)if(prime_flag[i] && !(gcdnum%(i+1))){
    ans++;
    while(!gcdnum%(i+1))gcdnum/=(i+1);
    }
    if(gcdnum > 1000000 && gcdnum != 1)ans++;
    printf("%lld",ans);
    return 0;
}