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) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        
        // a -> a'
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt() / 2;
        }
        
        int t = f(a[0]);
        
        // a' -> a''
        for (int i = 0; i < n; i++) {
            if (f(a[i]) != t) {
                System.out.println(0);
                return;
            }
            a[i] >>= t;
        }
        
        
        long l = 1;
        for (int i = 0; i < n; i++) {
            l = lcm(l, (long)a[i]);
            if (l > m) {
                System.out.println(0);
                return;
            }
        }
        
        int count = ((m / (int)l) + 1) / 2;
        
        System.out.println(count);
    }
    
    public static long lcm(long m, long n) {
        return m * (n / gcd(m, n));
    }
    
    public static long gcd(long a, long b) {
        if (b == 0) return a;
        
        return gcd(b, a % b);
    }
    
    public static int f(int x) {
        int res = 0;
        while (x % 2 == 0) {
            x /= 2;
            res++;
        }
        
        return res;
    }
}