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 9.2.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

#define rep(i,n) for(int i = 0;i<n;i++)
#define vv(Type,Name, n,m,d)vector<vector<Type>> vv(n, vector<Type>(m, d));

using namespace std;
using ll = long long;

ll gcd(ll a, ll b)
{
	if (a % b == 0)
	{
		return(b);
	}
	else
	{
		return(gcd(b, a % b));
	}
}

ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}


int main(void) {

	ios::sync_with_stdio(false);
	cin.tie(0);

	ll n, m;	cin >> n >> m;
	vector<ll> a(n);

	rep(i, n) {
		ll ii; cin >> ii;
		if (ii % 2 == 1) {
			cout << 0 << endl;
			return 0;
		}
		a[i] = ii / 2;
	}


	//cnt 2;
	int cnt = -1;
	rep(i, n) {
		int c = 0;
		ll b = a[i];
		while (true) {
			if (b % 2 == 1) break;
			b /= 2;
			c++;
		}

		if (cnt == -1) cnt = c;
		else if (cnt != c) {
			cout << 0 << endl;
			return 0;
		}
	}

	ll lc = 1;
	rep(i, n) lc = lcm(lc, a[i]);


	ll ans = 0;
	for (ll i = lc; i <= m; i += lc * 2) ans++;
	cout << ans << endl;





	return 0;
}