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 <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <complex>
#include <vector>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <algorithm>
#include <numeric>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define FOR(i, a, b) for(ll i = static_cast<ll>(a); i < static_cast<ll>(b); i++)
#define FORR(i, a, b) for(ll i = static_cast<ll>(a); i >= static_cast<ll>(b); i--)
#define REP(i, n) for(ll i = 0ll; i < static_cast<ll>(n); i++)
#define REPR(i, n) for(ll i = static_cast<ll>(n); i >= 0ll; i--)
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> IP;
typedef pair<ll, LP> LLP;

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};

constexpr int INF = 100000000;
constexpr ll LINF = 10000000000000000ll;
constexpr int MOD = static_cast<int>(1e9 + 7);
constexpr double EPS = 1e-9;

static inline ll mod(ll x, ll m)
{
    ll y = x % m;
    return (y >= 0 ? y : y+m);
}

// }}}

ll a, b;

map<ll, ll> prime_factor(ll n)
{
    map<ll, ll> ret;
    for(ll i = 2; i*i <= n; i++){
        while(n % i == 0){
            ret[i]++;
            n /= i;
        }
    }
    if(n != 1) ret[n]++;
    return ret;
}

void solve()
{
    map<ll, ll> pa, pb;
    pa = prime_factor(a);
    pb = prime_factor(b);
    ll ans = 0;
    ll n = max(a, b);
    for(ll i = 2; i*i <= n; i++){
        if(pa.find(i) != pa.end() && pb.find(i) != pb.end()){
            ans++;
        }
        if(i != n/i){
            if(pa.find(n/i) != pa.end() && pb.find(n/i) != pb.end()){
                ans++;
            }
        }
    }
    if(pa.find(n) != pa.end() && pb.find(n) != pb.end()){
        ans++;
    }
    cout << ans+1 << endl;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> a >> b;
    solve();
    return 0;
}

// vim:set foldmethod=marker: