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.Linq;

namespace ConsoleApp23 {
    class Program {
        static void Main(string[] args) {
            var s = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            long N = s[0];
            long M = s[1];

            long[] a = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            
            //aの全ての2の累乗の数が同じでなければいけない
            long a0factor = CountTwoFactor(a[0]);

            //Console.WriteLine(a0factor);

            for (int i = 0; i < N; i++) {
                if (CountTwoFactor(a[i]) != a0factor) {
                    Console.WriteLine(0);
                    return;
                }
            }

            //一度2のファクターで割った配列を用意する
            long[] a_divided = new long[N];
            for (int i = 0; i < N; i++) {
                a_divided[i] = a[i] / (long)Math.Pow(2, a0factor);
            }

            //Console.WriteLine(string.Join(" ", a_divided));

            //a_dividedの最小公倍数を求める
            long min_divided = a_divided[0];

            for (int i = 0; i < N; i++) {
                if (9223372036854775806 / min_divided < a_divided[i]) {
                    Console.WriteLine(0);
                    return;
                }

                min_divided = min_divided * a_divided[i] / Gcd(min_divided, a_divided[i]);
            }

            //Console.WriteLine(min_divided);

            //最初の一つ目の数
            if (9223372036854775806 / min_divided < (long)Math.Pow(2, a0factor - 1)) {
                Console.WriteLine(0);
                return;
            }

            long ans_first = min_divided * (long)Math.Pow(2, a0factor - 1);

            //Console.WriteLine(ans_first);

            //Mまでに、ans_first×奇数が何個あるか
            long ans = 0;
            long step = ans_first * 2;

            ans = (M - ans_first) / step + 1;

            //Console.WriteLine();

            Console.WriteLine(ans);
        }

        //最大公約数を求める
        public static long Gcd(long a, long 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;
        }
        
        static long CountTwoFactor(long A) {
            long cnt = 0;

            while (true) {
                if (A % 2 == 0) {
                    A /= 2;
                    cnt++;
                } else {
                    return cnt;
                }
            }
        }
    }
}