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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<ll, ll> Pair;
typedef vector<Pair> VP;

#define MOD 1000000007
#define times(n) rep(_idx, n)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define REP(i, s, n) for(int i = (int)(s); i < (int)(n); i++)
#define all(a) (a).begin(),(a).end()
#define pb push_back
#define mk make_pair
#define out(x) cout << (x) << endl;
#define DU(x) cerr << #x << " = " << (x) << endl;

template<typename _Ty>ostream& operator << (ostream& ostr, const vector<_Ty>& v) {if (v.empty()) {ostr << "[]";return ostr;}ostr << "[" << v.front();for (auto itr = ++v.begin(); itr != v.end(); itr++) {ostr << ", " << *itr;}ostr << "]";return ostr;}

VI primes(ll val) {
    VI v;
    ll tmp = val;
    REP(i, 2, sqrt(val) + 2) {
        if (tmp < i) { break; }
        if(tmp % i == 0) {
            v.pb(i);
            while(tmp % i == 0) tmp /= i;
        }
    }
    return v;
}

int main() {
    ll A, B;
    cin >> A >> B;
    VI a = primes(A), b = primes(B);

    sort(all(a)); sort(all(b));
    DU(a) DU(b)
    int ia = 0, ib = 0;
    int ans = 1;
    while(ia < a.size() && ib < b.size()) {
        if(a[ia] == b[ib]) {
            ans++;
            ia++; ib++;
        } else if (a[ia] < b[ib]) {
            ia++;
        } else {
            ib++;
        }
    }
    out(ans);
    return 0;
}