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 <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <stdbool.h>
typedef long long LL;
typedef long double Lf;
#define Rep(i,a,n)for(LL i=(LL)(a);i<(LL)(n);i++)
#define rep(i,n)Rep(i,0,n)
#define Repp(i,l,r,k)for(LL i=(l);i<(r);i+=(k))
#define Ma(x,y) ((x)>(y)?(x):(y))
#define mi(x,y) ((x)<(y)?(x):(y))
#define MOD 1000000007//1000000009,998244353 
int upi(const void* a, const void* b) { return*(int*)a<*(int*)b ? -1 : *(int*)a> * (int*)b ? 1 : 0; }
int downi(const void* a, const void* b) { return*(int*)a<*(int*)b ? 1 : *(int*)a> * (int*)b ? -1 : 0; }
void sortupi(int* a, int n) { qsort(a, n, sizeof(int), upi); } //1,2,3,4,...
void sortdowni(int* a, int n) { qsort(a, n, sizeof(int), downi); }
int upLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? -1 : *(LL*)a> * (LL*)b ? 1 : 0; }
int downLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? 1 : *(LL*)a> * (LL*)b ? -1 : 0; }
void sortupLL(LL* a, int n) { qsort(a, n, sizeof(LL), upLL); }
void sortdownLL(LL* a, int n) { qsort(a, n, sizeof(LL), downLL); }
int cmp_char(const void* a, const void* b) { return *(char*)a - *(char*)b; } // a,b,c,d
int cmp_str(const void* a, const void* b) { return strcmp(*(const char**)a, *(const char**)b); } // aaa,aab
LL gcd(LL p, LL q) { while (q) { LL t = p % q; p = q; q = t; }return p; }
LL modpow(LL a, LL n, int m) { LL x = 1; for (a %= m; n; n /= 2)n & 1 ? x = x * a % m : 0, a = a * a % m; return x; }
#define invmodpow(b,p)modpow(b,p-2,p)// a/b≡a*b^{-1}(mod p)...return b^{-1}(mod p)
typedef struct fac { LL n, exp; }F;
int fact(LL n,F*a){LL x,i,j,c=0;x=n;for(i=2;i*i<=x;i++){for(j=1;x%i==0;j++)x/=i;if(j>1){a[c].n=i;a[c].exp=j-1;c++;}}if(x-1){a[c].n=x;a[c].exp=1;c++;}return (int)c;}
//snippet
#define N 1009

int main(){
  LL a,b; F c[N];
  scanf("%lld%lld",&a,&b);
  printf("%d",fact(gcd(a,b),c)+1);
}