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"
using namespace std;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000
#define int long long

signed main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int A, B;
    cin >> A >> B;
    int ans = 1;
        
    REP(j, 2, 100000) {
        bool exist1 = false, exist2 = false;
        while(A % j == 0) {
            exist1 = true;
            A /= j;
        }
        while(B % j == 0) {
            exist2 = true;
            B /= j;
        }
        if (exist1 && exist2) ans++;
    }
    if (A == B && A != 1) ans++;
    cout << ans << endl;
    


    return 0;
}