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)

#![allow(dead_code)]
#![allow(non_snake_case)]

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

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

pub fn read_tuple<T: std::str::FromStr + Copy>() -> (T, T) {
    let v: Vec<T> = read_vec();
    (v[0], v[1])
}

pub fn gcd(a: i64, b: i64) -> i64 {
    if b == 0 {
        a
    } else {
        gcd(b, a % b)
    }
}

fn main() {
    let (N, M) = read_tuple::<i64>();
    let mut A: Vec<i64> = read_vec();
    for i in 0..N as usize {
        A[i] /= 2;
    }
    let lcm = A.iter().fold(A[0], |l, &a| l * a / gcd(l, a));
    for &a in A.iter() {
        if lcm / a % 2 == 0 {
            println!("0");
            return;
        }
    }
    let ans = (M / lcm + 1) / 2;
    println!("{}", ans);
}