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 <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <set>
#include <string>
#include <bitset>
#include <tuple>
#include <queue>
#include <map>
#include <cmath>
#include <iomanip>
using namespace std;
#define ll long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, k, n) for (int i = (int)(k); i <(int)(n); i++)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rREP(i,k,n) for (int i = (int)(n)-1; i >= k; i--)
#define INF 1LL<<61
ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
int main(){
    ll a,b;
    cin >> a >> b;
    ll c = gcd(a,b);
    ll ans = 1;
    for(ll i=2;i*i<=c;i++){
        if(c%i) continue;
        ans++;
        while(c%i==0)c/=i;
    }
    if(c!=1) ans++;
    cout << ans << endl;
    return 0;
}