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.*;
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[] ll = new long[n];
        boolean isOdd = false;
        boolean isEven = false;
        for(int i = 0; i < n; i++){
            long tmp = sc.nextLong()/2;
            ll[i] = tmp;
            if(tmp % 2 == 0){
                isEven = true;
            }else{
                isOdd = true;
            }
        }
        if(isEven && isOdd){
            System.out.println(0);
        }else if(isEven){
            System.out.println(solveByEven(ll,m));
        }else{
            System.out.println(solveByOdd(ll,m));
            
        }
        
    }
    
    private static long solveByEven(long[] ll, int m){
        long l = 1;
        for(int i = 0; i < ll.length; i++){
            if(l % ll[i] == 0 || ll[i] % l == 0){
                return 0;
            }
            l = lcm(l,ll[i]/2);
        }
        if(m < l){
            return 0;
        }else{
            m -= l;
            return 1+(m/(l*2));
        }
        
    }
    
    private static long solveByOdd(long[] ll, int m){
        long l = 1;
        for(int i = 0; i < ll.length; i++){
            l = lcm(l,ll[i]);
        }
        if(m < l){
            return 0;
        }else{
            m -= l;
            return 1+(m/(l*2));
        }
    }
    
    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));
    }
}