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: Accepted)

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <chrono>
#include <stack>
#include <string>
#include <bitset>
#include <unordered_map>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <cassert>
#include <tuple>
#include <iomanip>
#include <random>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <fstream>
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x.size())
#define pb push_back
#define mp make_pair
#define rnd mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define int long long
#define endl "\n"

using namespace std;

const int N = 500510;
const int inf = (int)(1e18) + 123;
const int mod = 998244353;
const double eps = 1e-9;

int gcd (int a, int b) {
    return b ? gcd (b, a % b) : a;
}

map <int, int> M;

signed main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, m; cin >> n >> m;
    int g = gcd(n, m);
    for(int i = 2; i * i <= g; ++i) {
        if (g % i == 0) {
            while (g % i == 0) {
                g /= i;
                M[i]++;
            }
        }
    }
    if (g != 1) {
        M[g]++;
    }
    cout << sz(M) + 1;

    return 0;
}