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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>

void prime_factorize(uint64_t x, uint64_t *n, uint64_t **p, uint64_t **q)
{
    uint64_t s;
    for(s=0; s*s<x; s++);
    *p = malloc(sizeof(uint64_t) * s);
    *q = malloc(sizeof(uint64_t) * s);

    int cps = 0;

    for(int i=2; i<=s; i++){
        int qi;
        for(qi=0; x%i==0; x/=i, qi++);
        if(qi!=0){
            (*p)[cps] = i;
            (*q)[cps] = qi;
            cps++;
        }
    }
    if(x!=1){
        (*p)[cps] = x;
        (*q)[cps] = 1;
        cps++;
    } 
    *n = cps;
}

uint64_t gcd(uint64_t x, uint64_t y)
{
    uint64_t z;
    while(y != 0){
        z = x % y;
        x = y;
        y = z;
    }
    return x;
}

uint64_t np(uint64_t x)
{
    uint64_t n, *p, *q;
    prime_factorize(x, &n, &p, &q);
    return n;
}

int main()
{
    uint64_t A, B;
    scanf("%"PRIu64"%"PRIu64, &A, &B);

    uint64_t c = gcd(A, B);

    printf("%"PRIu64"\n", np(c)+1);
    return 0;
}