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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp6
{
    class Program
    {
        static void Main(string[] args)
        {
            int [] NandM = Console.ReadLine().Split(' ').Select(c => int.Parse(c)).ToArray();
            int [] anums = Console.ReadLine().Split(' ').Select(c => int.Parse(c)).ToArray();

            int N = NandM[0];
            int M = NandM[1];

            //var num = Enumerable.Range(1, M).Where(x => IsHankoubai(anums, x)).Count();

            int lcm = Lcm(anums);

            if (lcm > M)
            {
                Console.WriteLine("0");
            }
            else
            {
                Console.WriteLine(M / Lcm(anums) + 1);
            }
        }

        public static bool IsHankoubai(IEnumerable<int> nums, int x) => nums.All(num => is_plusHalf((double)x / num));

        public static bool is_plusHalf(double num) => num % 1.0 == 0.5;

        // 最小公倍数
        public static int Lcm(int a, int b)
        {
            return a * b / Gcd(a, b);
        }

        public static int Lcm(IEnumerable<int> nums)
        {
            return nums.Aggregate((x, y) => Lcm(x, y));
        }

        // ユークリッドの互除法 
        public static int Gcd(int a, int b)
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }
    }
}