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: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<time.h>
#include<random>

using namespace std;

#define REP(i, n) for(int i = 0;i < n;++i)
#define REPR(i, n) for(int i = n-1;i >= 0;--i)
#define FOR(i, m, n) for(int i = m;i < n;++i)
#define FORR(i, m, n) for(int i = m-1;i >= n;--i)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define pb(a) push_back
#define print(x) cout<<(x)<<endl
#define pe(x) cout<<(x)<<" "
#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define lb(v,n) lower_bound(v.begin(), v.end(), n)
#define ub(v,n) upper_bound(v.begin(), v.end(), n)
#define int long long
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i,v.size())cout << v[i] << ((i == v.size() - 1) ? "\n" : " ")
template<typename T1, typename T2> inline void chmin(T1 & a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
typedef pair<int, int>P;
const int MOD = 998244353; const int MAX = 200020;
const double pi = acos(-1); const double EPS = 1e-12;
const ll INF = 2e18;

long long gcd(long long x, long long y) {
	long long m = max(x, y), n = min(x, y);
	if (m%n == 0)return n;
	else return gcd(m%n, n);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}
void solve() {
	int N, M; cin >> N >> M;
	vector<int>A(N);
	REP(i, N) {
		cin >> A[i];
		A[i] /= 2;
	}
	int G = A[0];
	REP(i, N) {
		G = lcm(G, A[i]);
		if (G > M) {
			print(0); return;
		}
	}
	REP(i, N) {
		if ((G / A[i]) % 2 == 0) {
			print(0); return;
		}
	}
	//DEBUG(G);
	if (G > M) {
		print(0); return;
	}
	int mx = (M - G) / (2 * G);
	print(max((ll)0, mx + 1));
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//int q;cin>>q;
	//while(q--)
	solve();
}