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 5.4.1) (judge verdict: Time Limit Exceeded)

// 途中でintをすべてlong long に変えたので ll と long long が混在
#include <stdio.h>

typedef long long ll;

void solve_142D();

long long GetGDC_ll(ll a, ll b); // 引数が long long な 最大公約数取得関数
long long GetNumFactors(long long n); // 素因数の数を数える

ll A, B; // 入力の２数字, 制約：A, B <= 10^12 なので long long

void solve_142D()
{
  long long gdc = GetGDC_ll(A, B);
  long long numFactors = GetNumFactors(gdc);
  printf("%lld\n", numFactors + 1);
}

long long GetGDC_ll(ll a, ll b)
// 引数が long long な 最大公約数取得関数
// 制約：a, b <= 10^12 なので GDC <= 10^6（で合ってるかな？）；戻り値は long long
{
  ll swap_tmp;
  if (a < b){ // 下の計算のために a > b にしておく
    swap_tmp = a; a = b; b = swap_tmp;
  }

  // ユークリッドの互除法により GDC を求める
  ll remainder = a % b;
  while (remainder != 0L){
    a = b; b = remainder; remainder = a % b;
  }
  return (long long)b;
}

long long GetNumFactors(long long n)
// 素因数の数を数える
// 戻り値：素因数の数
// 内容や何乗かといった情報は返さない
{
  if (n == 1) return 0; // GDC が 1 の場合は 0 を返して終了

  long long numFactor = 0;
  if (n % 2 == 0){ // 偶数の場合は numFactor を 1 にして 2 で割れるだけ割る
    numFactor = 1;
    while (1){
      n /= 2;
      if (n % 2 != 0) break;
    }
  }

  for (long long divisor = 3; divisor <= n; divisor += 2){
    if (n % divisor == 0){ // divisor で割れる場合は numFactor を +1 して divisor で割れるだけ割る
      ++numFactor;
      while (1){
        n /= divisor;
        if (n % divisor != 0) break;
      }
    }
  }

  return numFactor;
}

int main()
{
  scanf("%lld %lld", &A, &B);

  solve_142D();

  return 0;
}