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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Text;
using System.Numerics;
using System.IO;

namespace SolutionCS
{
    class Program
    {
        static bool AllEven(long[] val)
        {
            foreach(long v in val)
            {
                if(0 != v % 2)
                {
                    return false;
                }
            }

            return true;
        }

        static bool AllOdd(long[] val)
        {
            foreach (long v in val)
            {
                if (0 == v % 2)
                {
                    return false;
                }
            }

            return true;
        }

        // 最大公約数を求める
        // a=0の場合、bが結果として求まる
        // b=0の場合、aが結果として求まる
        static long gcd(long a, long b)
        {
            if (a < b)
            {
                return gcd(b, a);
            }

            while (b != 0)
            {
                long rem = a % b;
                a = b;
                b = rem;
            }

            return a;
        }

        // 最小公倍数、未検証
        static long lcm(long a, long b)
        {
            return a / gcd(a, b) * b;
        }

        static void Main(string[] args)
        {
            string[] s = Console.ReadLine().Split(' ');
            long N = long.Parse(s[0]);
            long M = long.Parse(s[1]);

            long[] val = new long[N];

            s = Console.ReadLine().Split(' ');
            for (long i = 0; i < N; i++)
            {
                val[i] = long.Parse(s[i]) / 2;
            }

            long scale = 1;

            while(AllEven(val))
            {
                for (long i = 0; i < N; i++)
                {
                    val[i] /= 2;
                }
                scale *= 2;
            }

            if(!AllOdd(val))
            {
                Console.WriteLine("{0}", 0);
                return;
            }


            long lcmv = 1;
            for (int i = 0; i < N; i++)
            {
                lcmv = lcm(lcmv, val[i]);
            }

            lcmv *= scale;

            M -= lcmv;
            if(M < 0)
            {
                Console.WriteLine("{0}", 0);
                return;
            }

            lcmv *= 2;

            Console.WriteLine("{0}", M / lcmv + 1);
            return;
        }
    }
}