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)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define ff first
#define ss second
#define mp make_pair
#define pp pair<ll,ll>
#define vi vector<pair<pp,pp> >
#define pq_min priority_queue<ll,vector<ll>,greater<ll>>
#define pq_max priority_queue<ll>
#define forx(i,a,b) for(ll i=0;i<a;i++)
#define ld long double
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef pair<pp,pp> ppll;
const ll mod=1e9 + 7;
vector<string> ans;


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

// Returns LCM of array elements 
ll findlcm(ll arr[], ll n) 
{ 
	// Initialize result 
	ll ans = arr[0]; 

	// ans contains LCM of arr[0], ..arr[i] 
	// after i'th iteration, 
	for (ll i = 1; i < n; i++) 
		ans = (((arr[i] * ans)) / 
				(gcd(arr[i], ans))); 

	return ans; 
} 


int main()
{
	fast;
    ll n,m;
    cin>>n>>m;
    ll a[n];
    map<ll,ll> s;
    for(ll i=0;i<n;i++)
    {
    	cin>>a[i];
    	s[a[i]]++;
    }
    ll v=findlcm(a,n);
    if(m < v)
    {
    	cout<<"0";
    	return 0;
    }
    m = m - v/2;
    ll u=ceil((ld)m/(ld)v);
    cout<<u<<endl;
}