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<iostream>
#include<stack>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<algorithm>
#include<iomanip>
#include<set>
#include<bitset>
#include<cmath>
#include<numeric>
#include<limits>
#include<functional>
#include <bitset>

#define REP(i, m, n) for(int i = (m); i < (n); i++)
#define rep(i,n) REP(i,0,n)
#define NIL -1
using namespace std;
typedef long long ll;
typedef long long int lli;
static const int MOD = 10000007;
static const int MAX = 100005;
// static const int NMAX = 5005;
static const lli INFTY = 1000000000000;

// typedef priority_queue<int, greater<int>> RevQ;
int N, K;
lli A, B;
// bool isprime[INFTY];
vector<lli> prime;
// vector<pair<int, int>> a(MAX);

lli gcd(lli a, lli b) {
  return b ? gcd(b, a%b) : a;
}
//
void eratho(lli n, bool isprime[]) {
  for (lli i = 0; i <= n; i++) {
    isprime[i] = true;
  }
  isprime[0] = isprime[1] = false;
  for (lli i = 2; i <= sqrt(n); i++) {
    if (isprime[i]) {
      lli j = i + i;
      while (j <= n) {
        isprime[j] = false;
        j += i;
      }
      // cout << "DETA" << endl;
    }
  }
}

void enterp(lli n, bool isprime[]) {
  eratho(n, isprime);
  for(lli i = 2; i <= n; i++){
    if (isprime[i]) {
      prime.push_back(i);
    }
  }
}


int main(){
  cin >> A >> B;
  // int cnt = 0;
  lli d;
  // cout << "HEY!" << endl;
  if (A < B) swap(A,B);
  d = gcd(A, B);
  lli D = d;
  int cnt = 1;
  lli i = 2;
  if (i <= d) {
    int flag = 0;
    while (d % i == 0) {
      d /= i;
    }
    if (flag) cnt++;
  }
  i = 3;
  while (i <= d) {
    int flag = 0;
    while(d % i == 0) {
      flag = 1;
      d /= i;
    }
    if (flag) cnt++;
    i += 2;
  }
  cout << cnt << endl;
}