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)

use std::io::{self, BufRead};

fn main() {
    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let n_m = lines.next().unwrap().unwrap();
    let mut n_m = n_m.split_whitespace();
    let n: i64 = n_m.next().unwrap().parse().unwrap();
    let m: i64 = n_m.next().unwrap().parse().unwrap();
    let a = lines.next().unwrap().unwrap();
    let a: Vec<i64> = a.split_whitespace().map(|s| s.parse().unwrap()).collect();

    let mut even = false;
    let mut odd = false;

    let mut multiple = 1;
    for &num in a.iter() {
        let num = num / 2;
        if num % 2 == 0 {
            even = true;
        } else {
            odd = true;
        }
        if even && odd {
            println!("0");
            return;
        }
        multiple = lcm(multiple, num);
        if multiple > m || multiple / num % 2 == 0 {
            println!("0");
            return;
        }
    }

    let ans = (m + multiple) / (2 * multiple);
    println!("{}", ans);
}

fn gcd(a: i64, b: i64) -> i64 {
    let mut t = (a, b);
    while t.1 != 0 {
        t = (t.1, t.0 % t.1);
    }
    t.0
}

fn lcm(a: i64, b: i64) -> i64 {
    a / gcd(a, b) * b
}