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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b)for(int i=(a);i<=(b);i++)
#define repR(i,n) for(int i=n;i>=0;i--)
#define P pair<int, int>
#define sz(x) x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<int>
#define COUT(x) cout<<(x)<<endl
#define MA map<int,int>
#define SE set<int>
#define PQ priority_queue<int>
#define PQR priority_queue<int,VE,greater<int>>
#define COUT(x) cout<<(x)<<endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
	if (a.second != b.second) {
		return a.second < b.second;
	}
	else {
		return a.first < b.first;
	}
}
const long long MOD = 1000000007;
const long long INF = 1LL << 50;
// pair -> vector<pair<int,int>> 
//		   AandB.push_back(make_pair(a[i],b[i]))

signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int a, b; cin >> a >> b;
	set<int> ko;
	int now = 2;
	int tmp = 1;
	while (true) {
		if (gcd(a/tmp, b/tmp) < now)break;
		
			
		if(prime(now))ko.insert(now);
		tmp *= now;
		now++;
	}
	cout << ko.size() + 1 << endl;
	return 0;
}