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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        // write your code here
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        long m = sc.nextLong();
        ArrayList<Long> list = new ArrayList<>();
        long gcd = sc.nextLong() / 2;
        list.add(gcd);

        long ni = 1;

        long niValue = gcd;
        while (niValue % 2 == 0) {
            ni *= 2;
            niValue /= 2;
        }
        long ni2 = ni * 2;
        for (int i = 1; i < n; i++) {
            long value = sc.nextLong() / 2;
            if (list.contains(value)) {
                continue;
            }
            // 2の個数チェック
            if (value % ni != 0 || value % ni2 == 0) {
                System.out.println(0);
                return;
            }
            list.add(value);
            gcd = gcd(gcd, value);
        }

        double lcm = 1;
        double standard = (double)1 / m;
        for (Long aLong : list) {
            lcm *= 1.0 / aLong;
            if (lcm * gcd < standard) {
                System.out.println(0);
                return;
            }
        }

        lcm *= gcd;
        double ans = Math.floor(m * lcm);

        System.out.println((long)Math.ceil(ans / 2));
    }

    private static long gcd(long a, long b) {
        long temp;
        while ((temp = a % b) != 0) {
            a = b;
            b = temp;
        }
        return b;
    }
}