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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <math.h>
#include <vector>
#include <queue>
#include <iomanip>
#include <string>
#include <algorithm>
using namespace std;
using Graph = vector<vector<int>>;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep_1 (i,j,n) for(int i=j;i<n;i++)
typedef long long ll;
typedef long l;
const ll l_INF = 1010101010101010;
const int INF = 1010101010;
const int dx[4] = {-1,0,1,0};
const int dy[4] = {0,-1,0,1};
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int main(){
  ll a,b; cin>>a>>b;
  if(a>b) swap(a,b);
  while(b%a!=0){
    b=b%a;
    if(a>b) swap(a,b);
    if(a==0) break;
  }
  //cout<<sqrt(a)<<endl;
  int cnt=1;
  ll c=a;
  for(int i=2;i<=sqrt(a);i++){
    bool judge=false;
    while(a%i==0){
      a/=i;
      judge=true;
    }
    if(judge){
      cnt++;
      //cout<<i<<endl;
    }
  }
  if(a>1) cnt++;
  cout<<cnt<<endl;
}