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: Accepted)

import java.util.*;
import java.io.*;
import java.math.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long l = 1;
        long prev = 0;
        for(int i = 0; i < n; i++){
            long temp = sc.nextLong()/2;
            l = lcm(l,temp);
            if(i == 0){
                prev = divide2(temp);
            }else{
                if(prev != divide2(temp)){
                    System.out.println(0);
                    return;
                }
            }
        }
        if(m < l){
            System.out.println(0);
        }else{
            if((m/l) % 2 == 0){
                System.out.println(m/(l*2));
            }else{
                System.out.println(1+m/(l*2));
            }
        }
        
    }

    private static long divide2(long l){
        long ret = 0;
        while(l % 2 == 0){
            ret++;
            l /= 2;
        }
        return ret;
    }
    
    private static long gcd(long a, long b)
    {
        while (b > 0)
        {
            long temp = b;
            b = a % b; // % is remainder
            a = temp;
        }
        return a;
    }
    
    private static long lcm(long a, long b)
    {
        return a * (b / gcd(a, b));
    }
}