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: Runtime Error)

import java.util.*;

public class Main{
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        long m = Long.parseLong(sc.next());
        
        long[] a = new long[n];
        for(int i=0; i<n; i++){
            a[i] = Long.parseLong(sc.next());
        }
        
        int h2 = 0;
        {
            long tmp = a[0];
            while(tmp%2==0){
                tmp /= 2;
                h2++;
            }
        }
        
        long fact = 1;
        for(int i=0; i<n; i++){
            int cnt2 = 0;
            {
                long tmp = a[i];
                while(tmp%2==0){
                    tmp /= 2;
                    cnt2++;
                }
            }
            if(cnt2 != h2){
                System.out.println(0);
                return;
            }
            fact = lcm(fact, a[i]);
        }
        
        System.out.println((m+fact)/fact);
    }
    
    public static long lcm(long m, long n) {
        return m * n / gcd(m, n);
    }
    
    public static long gcd(long m, long n) {
        if(m < n) return gcd(n, m);
        if(n == 0) return m;
        return gcd(n, m % n);
    }
}