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)

#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout <<"DEBUG"<<" "<< #x << ":" << x << '\n'

//↓0-originか1-originでn回繰り返し
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   //0-indexed昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) //1-indexed昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  //0-indexed降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   //1-indexed降順
#define all(x) (x).begin(),(x).end()

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii; //2次元配列
typedef vector<ll> vll;
typedef pair<int, int> PII;

template<typename T>
using vec = vector<T>;

template<class T,class U> inline bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U> inline bool chmin(T &a, const U &b){if(a>b){a=b;return 1;}return 0;}

const ld PI = acos(-1.0L);
const ll infll = (1LL<<62)-1;
const int inf = (1<<30)-1;
const int mod = 1000000007;
//const int mod = 998244353;

/*FUNCs=================================================*/
long long GCD(long long a, long long b){
    return (b ? GCD(b, a%b) : a);
}

// LCM == a*b/GCD(a,b)
/*MAIN==================================================*/
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化
    //cout << fixed << setprecision(10);

    ll n, m; cin >> n >> m;
    vll a(n);
    rep(i,n) {
        cin >> a[i];
        a[i]/=2;
    }
    ll lcm = a[0];
    for(auto i : a) {
        lcm = lcm*i/GCD(lcm, i);
    }
    for(auto i : a) {
        if(lcm/i%2==0) {
            cout << 0 << '\n';
            return 0;
        }
    }
    cout << (m / lcm + 1)/ 2 <<'\n';
}