Problem:
Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io;
use std::str::FromStr;
use std::collections::HashSet;

fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn input_vec() -> Vec<u64> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
      a
    } else {
      gcd(b, a % b)
    }
}

fn dru(n: u64, m: u64) -> u64 {
    if n % m == 0 {
        return n / m;
    } 
    n / m + 1
}

fn main() {
    let (n, m) = input_tuple();
    let mut a_vec = input_vec();
    for a in &mut a_vec {
        *a /= 2;
    }
    let mut base_lcm = 1;
    for i in 1..a_vec.len() {
        let lcm = a_vec[0] * a_vec[i] / gcd(a_vec[0], a_vec[i]);
        if lcm / a_vec[0] % 2 == 1 && lcm / a_vec[i] % 2 == 1 {
            base_lcm = base_lcm * lcm / gcd(base_lcm, lcm);            
        } else {
            println!("0");
            return;
        }
    }
    if base_lcm > m {
        println!("0");
    }
    println!("{}", dru(m / base_lcm, 2));
}