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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define ld double
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> order_set;
const ll mod = 1e9 + 7, p = 115211;
const ll N = 2.5e3 + 5, M = 101;
const ld pi = acos(-1);

ll power(ll x, unsigned ll y, ll p)
{
     ll res = 1; 
     x = x % p;
     while (y > 0)
     {
         if (y & 1)
             res = (res*x) % p;
         y = y>>1;
         x = (x*x) % p;  
     }
    return res;
}


int main() 
{ 
    FAST;
   // freopen("input.in","r",stdin);
   // freopen("output.out","w",stdout);
    ll a, b;
    cin >> a >> b;
    set <ll> s1, s2;
    for(ll i=2;i<=sqrt(a);i++) {
        if(a % i == 0) {
            s1.insert(i);
            while(a % i == 0) {
                a /= i;
            }
        }
    }
    if(a != 1) {
        s1.insert(a);
    }
    for(ll i=2;i<=sqrt(b);i++) {
        if(b % i == 0) {
            s2.insert(i);
            while(b % i == 0) {
                b /= i;
            }
        }
    }
    if(b != 1) {
        s2.insert(b);
    }
    ll c = 1;
    for(auto it=s1.begin();it!=s1.end();it++) {
        if(s2.count(*it) == 1) {
            c++;
        }
    }
    cout << c;
    return 0;
}