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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;
typedef pair<lint, int> pli;
typedef complex<double> xy_t;

const lint mod = 1e9 + 7;

lint gcd(lint p, lint q){
    if(p < q) swap(p, q);
    while(q){
        lint r = p % q;
        p = q;
        q = r;
    }
    return p;
}

bool is_prime(lint n){
    for(lint i=2; i*i<=n; ++i){
        if(n % i == 0) return false;
    }
    return true;
}

vector<lint> getdiv(lint n){
    vector<lint> ret;
    for(lint i=1; i*i<=n; ++i){
        if(n % i == 0){
            ret.push_back(i);
            if(i*i != n) ret.push_back(n / i);
        }
    }
    return ret;
}

const int MAX = 1000010;
bool sieve[MAX];

void getsieve(){
    rep(i, MAX) sieve[i] = true;
    for(int i=2; i<MAX; ++i)if(sieve[i]){
        for(int j=i*2; j<MAX; j+=i) sieve[j] = false;
    }
    return;
}

int main(){
    lint a, b;
    scanf("%lld%lld", &a, &b);
    lint g = gcd(a, b);
    auto d = getdiv(g);
    sort(d.begin(), d.end());
    d.erase(unique(d.begin(), d.end()), d.end());
    getsieve();

    int ans = 0;
    for(int n: d)if((n < MAX && sieve[n]) || is_prime(n)) ++ans;
    printf("%d\n", ans);
}