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 <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define all(a) a.begin(),a.end()
#define F first
#define S second
#define pb push_back
#define ll long long
#define ld long double
#define mkp make_pair
#define sz(x) (int)x.size()
#define pi pair<long long, long long>
#define watch(x) cout << (#x) << " is " << (x) << endl; cout.flush()
#define endl '\n'
#define _ << ' '
#define FAST_LOG2(x) (sizeof(unsigned long)*8 - 1 - __builtin_clzl((unsigned long)(x)))
#define clog2(x) (((x) - (1 << FAST_LOG2(x))) ? FAST_LOG2(x) + 1 : FAST_LOG2(x))
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> idxset;
//find_by_order(i) : iterator to the value at index i, order_of_key(x) : the index of value x
const ll mod = (ll)1e9+7;
const ll INF = (ll)1e9+1;

int p[1000001];
ll a, b, ans=0;

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    #ifndef ONLINE_JUDGE
    //freopen("B.txt", "r", stdin);
    #endif
    for(int i=2; i<=500000; i++){
        if(p[i]) continue;
        for(int j=i*2; j<=1000000; j+=i){
            p[j] = 1;
        }
    }
    cin >> a >> b;
    for(ll i=1; i*i<=min(a, b); i++){
        if(a%i==0&&b%i==0){
            if(!p[i]) ans++;
        }
    }
    cout << ans;
}