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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

long long n, m;
long long a[100000];
double rs;

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	double x = 1;
	while (a[1]*(x+0.5)<=m)
	{
		bool check = true;
		float tmp = a[1] * (x + 0.5);
		for (int i = 2; i <= n; i++)
		{
			float t1= tmp / a[i];
			double t2= tmp / a[i];
			if (t1 - t2 != 0.5)
			{
				check = false;
				break;
			}
		}
		if (check) { rs++; }
		x++;
	}
	cout << rs;
	return 0;
}