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) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] a = new int[N];
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
        }

        Arrays.sort(a);
        int gcd = a[0];
        for (int i = 1; i < N; i++) {
            if (a[i]%2 == 1) {
                System.out.println(0);
                return;
            } 
            gcd = calcGcd(gcd, a[i]/2);
        }

        boolean ok = true;
        int xx = f(a[0]);
        for (int i = 1; i < N; i++) {
            if (f(a[i]) != xx) {
                ok = false;
                break;
            }
        }
        if (!ok) {
            System.out.println(0);
            return;
        }

        long lcm = a[0]/2;
        for (int i = 1; i < N; i++) {
            if (a[i]/2 % lcm == 0) {
                lcm *= a[i]/2/lcm;
            } else {
                lcm *= a[i]/2/gcd;
            }
        }
        //System.out.println(lcm);

        System.out.println((M/lcm+1)/2);
    }

    private static int calcGcd(int a, int b) {
        if (b == 0) {
            return a;
        }

        return calcGcd(b, a%b);
    }

    private static int f(int x) {
        int ret = 0;
        while (x%2 == 0) {
            ret++;
            x /= 2;
        }

        return ret;
    }
}