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>
#define REP(i, n) for(int i = 0; i < n; ++i)
#define REPP(i, n) for(int i = 1; i <= n; ++i)
#define REPR(i, n) for(int i = n-1; i >= 0; --i)
#define FOR(i, m, n) for(int i = m; i < n; ++i)
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b) {if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b) {if(a<b) a=b;}

template<typename T>
map<T, T> primeFactor(T n) {
    map<T,T> res;
    for(T i = 2; i*i<=n; ++i) {
        while(n%i==0) {
            res[i] ++;
            n /= i;
        }
    }
    if(n != 1) res[n] = 1;
    return res;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll a,b; cin >> a >> b;
    map<int,int> m;
    for(auto x : primeFactor(a)) {
        ++ m[x.first];
    }
    for(auto x : primeFactor(b)) {
        ++ m[x.first];
    }
    int ans = 1;
    for(auto x : m) {
        if(x.second > 1) {
            ++ ans;
        }
    }
    cout << ans << endl;
    return 0;
}