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::*;

fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
        return a;
    }
    gcd(b, a % b)
}
fn lcm(a: u64, b: u64) -> u64 {
    a / gcd(a, b) * b
}
fn gcd_array(a: &Vec<u64>) -> u64 {
    let mut res = a[0];
    for i in 1..a.len() {
        res = gcd(res, a[i]);
    }
    res
}
fn lcm_array(a: &Vec<u64>) -> u64 {
    let mut res = a[0];
    for i in 1..a.len() {
        res = lcm(res, a[i]);
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: u64 = itr.next().unwrap().parse().unwrap();
    let a: Vec<u64> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let l = lcm_array(&a);
    let g = gcd_array(&a);

    for i in 0..n {
        if a[i] / g % 2 == 0 {
            println!("0");
            return;
        }
    }

    // a_k / 2 * (2 * p + 1) = lcm/2 * i (lcm * i <= M)
    // あるkに関して、仮にa_kとlcmの2の約数の数が違うなら、2 * p + 1が奇数であるため、そのようなpは存在しない。
    // 存在する場合、lcm / 2の奇数倍を求めれば良い。

    println!("{}", (m + l - 1) / l);
}