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: Accepted)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

fn two_count(n: u64) -> u32 {
    let mut n = n;
    let mut res = 0;
    while n % 2 == 0 {
        res += 1;
        n /= 2;
    }
    res
}

fn solve() {
    input! {
        n: usize, m: u64,
        av: [u64; n]
    };

    let two_counts = av.iter().map(|&a| two_count(a)).collect::<Vec<_>>();
    let is_ok = two_counts.iter().collect::<HashSet<_>>().len() == 1;
    if !is_ok {
        println!("0");
        return;
    }
    let two_p = two_counts[0];
    let odd_lcm = av
        .iter()
        .map(|&a| {
            let m = 2.pow(two_p);
            a / m
        })
        .fold(1, |acc, x| acc.lcm(&x));
    let first_gcd = 2_u64.pow(two_p - 1) * odd_lcm;
    let res = if first_gcd > m {
        0
    } else {
        1 + (m - first_gcd) / (first_gcd * 2)
    };
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}