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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

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

fn lcm(a: usize, b: usize) -> usize {
    a * b / gcd(a, b)
}

fn main() {
    let input = read_vec::<usize>();
    let _n = input[0];
    let m = input[1];
    let mut an = read_vec::<usize>();

    an = an.iter().map(|a| a / 2).collect();
    // println!("{:?}", an);

    let l = an.iter().fold(1, |acc, &x| lcm(acc, x));

    println!("{}", (m+l) / l / 2);

    // let mut count = 0;

    // for i in 0..=m {
    //     let x = l * (2 * i + 1);
    //     if x > m {
    //         break;
    //     }
    //     // println!("{}", x);
    //     count += 1;
    // }

    // println!("{}", count);


}