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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn gcd(a: u64, b: u64) -> u64 {
    let mut c = cmp::max(a, b);
    let mut d = cmp::min(a, b);

    while c%d > 0 {
        let tmp = d;
        d = c%d;
        c = tmp;
    }

    d
}


fn main() {
    let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = line[0];
    let m = line[1] as u64;
    let a: Vec<u64> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();

    let mut tmp = 1;

    for i in 0..n {
        tmp = tmp * (a[i]/2) / gcd(tmp, a[i]/2);
    }

    let ans = if tmp <= m {
        1 + (m - tmp) / (tmp*2)
    } else {
        0
    };

    println!("{}", ans);
}