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;using f64 = double;using i64=long long;using u64=unsigned long long;
template<typename T> using Arr=vector<T>;
#define hfor(v, s, e) for(int v=(s);(s)<=v&&v<(e);++v)//half-opened range
#define hfori(v, s, e) for(int v=(e)-1;(s)<=v&&v<(e);--v)//inversion
#define hforo(v, s, e) int v=(s);for(;(s)<=v&&v<(e);++v)//out declaration
#define hforoi(v, s, e) int v=(e)-1; for(;(s)<=v&&v<(e);--v)
#define cfor(v, s, e) hfor(v,(s),(e)+1)//closed range
#define cfori(v, s, e) hfori(v,(s),(e)+1)
#define cforo(v, s, e) hforo(v,(s),(e)+1)
#define cforoi(v, s, e) hforoi(v,(s),(e)+1)
#define rep(v,x) hfor(v,0,(x))
#define repi(v,x) hfori(v,0,(x))
#define repo(v,x) hforo(v,0,(x))
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pushb push_back
#define pushf push_front
#define popb pop_back
#define popf pop_front
#define empl emplace
#define emplb emplace_back
#define emplf emplace_front
#define fi first
#define se second
#define cxp constexpr
#define PQ std::priority_queue

#ifndef DEBUG
	#pragma GCC optimize ("Ofast")
	auto __PRE_RUN__=(ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),(cout<<fixed<<setprecision(11)), 0);
#endif

template<typename T> cxp T inf() { return numeric_limits<T>::max() / 2; }
auto mapf(auto a, auto f){for(auto& x:a)x=f(x); return a;}
int rd(int lb, int ub){static mt19937 rng(time(0)^i64(new int)); return uniform_int_distribution<int>(lb, ub-1)(rng);}
int rd(int ub=inf<int>()){return rd(0,ub);}
const f64 pi=acosl(-1);
#define endl '\n'//not interactive?
#define int i64//MLE?
using pint = pair<int,int>;
using tint = tuple<int,int,int>;

int gcd(int a, int b){
	if(!b)
		return a;
	return gcd(b, a%b);
}
int lcm(int a, int b){return a/gcd(a,b)*b;}

signed main(){
	int n, m; cin>>n>>m;
	int a[n];
	int z=1;
	rep(i,n){
		cin>>a[i];
		if(z>m)
			break;
		z=lcm(z,a[i]/2);
	}
	cout<<m/z/2+m/z%2<<endl;

	return 0;
}