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>
using namespace std;
using ll=long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
#define all(x) x.begin(),x.end()
#define rep(i,j,n) for (long long i = j; i < (long long)(n); i++)
#define _GLIBCXX_DEBUG
const ll MOD = 1000000007;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//(a+b-1)/b
//priority_queue<ll, vector<ll>, greater<ll>> q;
ll dp[100100100];

signed main(){
	//cout << fixed << setprecision(10);
	ll n,m; cin>>n>>m;
	ll sum_gcd=1;
	bool ok=false;
	rep(i,0,n){
		ll a; cin>>a;
		a/=2;
		sum_gcd=(sum_gcd*a)/__gcd(sum_gcd,a);
		if(sum_gcd>m){
			ok=1;
			break;
		}

	}
	if(ok){
		cout << 0<< endl;
		return 0;
	}
	ll ans = m/sum_gcd/2;
	if((m/sum_gcd%2==1))ans++;
	cout << ans << endl;
	


	return 0;
}