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;
        }
    }
    // X = a_i * (p + 0.5)
    // X = (a_i / 2) * (p * 2 + 1)
    // あるaについて、Xにするのに足りない数をpで補うことができれば、Xは半公倍数となる。
    // 最小公倍数を考える。これをXとした時、あるaをXにするのに足りない数をpで補える。
    // しかし、p * 2 + 1は奇数なので、足りない数が偶数の時、Xを作れない。
    // Xは最小公倍数なので、そのような時、結局一つも作れない。
    // 答えは(最小公倍数 / 2)の奇数倍
    println!("{}", (m + l - 1) / l);
}