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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma region
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue> 
#include <stack>
#include <set>
#include <list>

using namespace std;
typedef long long ll;
//#define rep(i, s, e) for (int(i) = (s); (i) < (e); ++(i))
#define rep(i, e) for (int(i) = 0; (i) < (e); ++(i))
#define all(x) x.begin(),x.end()
#pragma endregion

#pragma region GCD
int gcd(ll a, ll b)
{
	if (a%b == 0)return b;
	return gcd(b, a%b);
}
#pragma endregion
#pragma region LCM
int lcm(ll a, ll b)
{
	return (a*b) / gcd(a, b);
}
#pragma endregion

int main()
{
	int n; cin >> n;
	ll m; cin >> m;
	vector<ll> a(n); rep(i, n)cin >> a[i];
	ll tmp = a[0];
	if ((a[0] / 2) % 2 == 1)
	{
		rep(i, n)
		{
			if ((a[i] / 2) % 2 == 0)
			{
				cout << 0 << endl;
				return 0;
			}
			tmp = lcm(tmp, a[i]);
		}
		cout << (m + tmp / 2) / tmp << endl;
	}
	else
	{
		rep(i, n)
		{
			if ((a[i] / 2) % 2 == 1 || (a[i] / 4) % 2 == 0)
			{
				cout << 0 << endl;
				return 0;
			}
			tmp = lcm(tmp, a[i]);
		}
		cout << (m + tmp / 2) / tmp << endl;
	}
}