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: Accepted)

fn gcd(x: usize, y: usize) -> usize {
    let mut x = x;
    let mut y = y;
    while y != 0 {
        let t = y;
        y = x % y;
        x = t;
    }
    x
}

fn lcm(x: usize, y: usize) -> usize {
    x * y / gcd(x, y)
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: usize = iter.next().unwrap().parse().unwrap();

    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let a_s = buf.split_whitespace()
        .map(|x| x.parse::<usize>().unwrap())
        .collect::<Vec<_>>();

    let mut div_cnt = 0usize;
    let mut a = a_s[0];
    loop {
        if a % 2 == 1 {
            break;
        }
        a /= 2;
        div_cnt += 1;
    }
    let div = 2usize.pow(div_cnt as u32);

    let mut ans = 0usize;
    if a_s.iter().find(|&&x| (x < div) || ((x/div)%2 == 0)) == None {
        let lcm_all = a_s.iter().fold(1, |lcm_all, a| lcm(lcm_all, (*a)/2));
        let lcm_mul = m / lcm_all;
        ans = lcm_mul / 2 + lcm_mul % 2;
    }

    println!("{}", ans);
}