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 implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int N = Integer.parseInt(scanner.next());
        long M = Integer.parseInt(scanner.next());
        long[] A = new long[N];
        long max = 0;
        for (int i = 0; i < N; i++) {
            A[i] = Integer.parseInt(scanner.next());
            max = Math.max(max, A[i] / 2);
        }
        long gcd = A[0] / 2;
        long lcm = A[0] / 2;
        Set<Long> used = new HashSet<>();
        used.add(A[0]);
        for (int i = 1; i < N; i++) {
            if (used.add(A[i])) {
                gcd = gcd(gcd, A[i] / 2);
                lcm = lcm * (A[i] / 2) / gcd;
                if (lcm > M) {
                    System.out.println(0);
                    return;
                }
            }
        }

        long ans = M / lcm;
        if (lcm == max) {
            ans++;
        }
        System.out.println(ans - ans / 2);
    }

    static long gcd(long x, long y) {
        if (x < y) {
            return gcd(y, x);
        }
        if (y == 0) {
            return x;
        }
        return gcd(y, x % y);
    }
}