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>

using namespace std;

#define fRead(x)        freopen(x,"r",stdin)
#define fWrite(x)       freopen (x,"w",stdout)

#define LL              long long
#define ULL             unsigned long long
#define ff              first
#define ss              second
#define pb              push_back
#define INF             2e16
#define PI              acos(-1.0)
#define mk              make_pair
#define pii             pair<int,int>
#define pll             pair<LL,LL>


#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define SQR(a)          ((a)*(a))
#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define MEM(a,x)        memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))

#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())


#define FastRead        ios_base::sync_with_stdio(0);cin.tie(nullptr);

int main()
{
    LL a,b;
    cin >> a >> b;
    LL g = __gcd(a,b);
    LL ans = 0;
    LL sq = sqrt(g);
    for(int i = 2;i <= sq;i++){
        if(g % i == 0){
            ans++;
            while(g % i == 0)g/=i;
        }
    }
    if(g > 1)ans++;
    cout << ans + 1 << "\n";
}