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: Accepted)

#include<bits/stdc++.h>
using namespace std;
//Optimisations
#pragma GCC target ("avx2")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("O2")
//shortcuts for functions
#define pb push_back
#define mp make_pair
#define ff first 
#define ss second
#define all(v) v.begin(),v.end()
#define prec(n) fixed<<setprecision(n)
#define n_l '\n'
// make it python 
#define gcd __gcd
#define append push_back
#define str to_string
// utility functions shortcuts
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define sswap(a,b) {a=a^b;b=a^b;a=a^b;}
#define swap(a,b) {auto temp=a; a=b; b=temp;}
#define init(dp) memset(dp,-1,sizeof(dp));
#define set0(dp) memset(dp,0,sizeof(dp));
#define bits(x) __builtin_popcount(x)
#define SORT(v) sort(all(v))
#define endl "\n"
#define forr(i,n) for(ll i=0;i<n;i++)
// declaration shortcuts
typedef long long int ll;
#define int ll
// Constants
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr ll INF = 1999999999999999997; 
constexpr int inf= INT_MAX;
constexpr int MAXSIZE = int(1e6)+5;
constexpr auto PI  = 3.14159265358979323846L;
constexpr auto oo  = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;
constexpr auto maxn = 100006;
//void IOfile(){
//freopen(file_name, reade_mode, stdin);
//freopen(file_name, write_mode, stdout);
//}
void fastio(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
int32_t main(){
	fastio();
	int n , m;
	cin >> n >> m;
	set<int> s;
	int lcm = 2;
	int a[n];
	forr(i,n) cin >> a[i];
	forr(i,n){
		int c = 0;

		int x = a[i];

		while(x % 2 == 0){
			x = x/2;
			c++;
		}
		s.insert(c);
		lcm = lcm * a[i]/__gcd(lcm,a[i]);
		if(s.size() > 1 || lcm > 2*m){
			cout << 0;
			return 0;
		}
	}
	cout << (2 * m / lcm + 1) /2;
}