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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

fn euk(a: usize, b: usize) -> usize {
   if b == 0 {
      a
   } else {
      euk(b, a % b)
   }
}

#[fastout]
fn main() {
   input! {
      a: usize,
      b: usize,
   }

   let mut m = euk(a, b);
   let mut primes = vec![1];
   for i in 2.. {
      if i * i > m {
         break;
      }
      if m % i == 0 {
         primes.push(i);
         while m % i == 0 {
            m /= i;
         }
      }
   }
   if m != 1 {
      primes.push(m);
   }
   let ans = primes.len();
   println!("{}", ans);
}