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 (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;


#[fastout]
fn main() {
    input! {
    n: usize,
    m: usize,
    mut a: [usize; n]
    }
    let a2 = a.iter().map(|x| x / 2).collect::<Vec<_>>();
    let mut l = a2[0];
    for i in  0..n - 1 {
        if div_two(a2[i]) != div_two(a2[i + 1]) {
            print!("0");
            return
        }
        l = l * a2[i+1] / gcd(&l, &a2[i + 1]);
    }
    let ans = if l % 2 == 1 {(m / l + 1) / 2} else {m / l / 2};
    println!("{}", ans);
}

// O(log(N))
fn gcd(a: &usize, b: &usize) -> usize {
    let (mut x, mut y) = if a < b { (*b, *a) } else { (*a, *b) };
    let mut r = x % y;
    while r != 0 {
        x = y;
        y = r;
        r = x % y;
    }
    y
}

fn div_two(n: usize) -> usize {
    let mut n = n;
    let mut res = 0;
    while n % 2 == 0 {
        n /= 2;
        res += 1;
    }
    res
}