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>
#include <cmath>
#include <numeric>
#include <iomanip>
// #include <regex> //正規表現
using namespace std;
#define rep(i,a,b) for(int i=(a); i<(b); ++i) // a ≦ i < b 
#define Rrep(i,a,b) for(int i=(a);i>=(b);--i) // reverse repeat. a から b まで減少．
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend() //逆イテレータ
#define RANGE(a,b,c) (a).begin()+b,(a).begin()+c // コンテナ a の 要素 b から c へのイテレータ
#define MOD 1000000007ll
#define INF 1000000000
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;
// グリッドグラフの移動ベクトル
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

int main() {
cin.tie(0);
ios::sync_with_stdio(false);

int64_t A, B; cin >> A >> B;

// 全ての素数同士は互いに素
// 素数をすべて選ぶ．
// エラトステネスのふるい作成．iが素数ならis_prime[i]=1．作成はO(NloglogN)．素数判定がO(1)でできる．
int MAX = (int)sqrt(max(A,B)) + 1;
vector<bool> is_prime(MAX, 1);
is_prime[0] = 0, is_prime[1] = 0;
for (int i = 2; i < MAX; ++i) {
    if (!is_prime[i]) continue;
    for (int j = i*2; j < MAX; j += i) is_prime[j] = 0;
}

// 公約数の集合
int cnt = 1;
for (int64_t i = 1; i * i <= min(A, B); ++i){ // min(A,B)以下の約数を探す．
    if (A % i == 0 && B % i == 0 && is_prime[i]) {
        ++cnt;
    }
}

cout << cnt << "\n";


}

// 境界，出力文字列 チェック