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: Runtime Error)

#include <bits/stdc++.h>
#define fastcin() cin.tie(0);ios::sync_with_stdio(false)
#define rep(i,n) for(int i=0;i<int(n);i++)
#define rep1(i,n) for(int i=1;i<=int(n);i++)
#define per(i,n) for(int i=int(n)-1;i>=0;i--)
#define per1(i,n) for(int i=int(n);i>0;i--)
#define foreach(i,n) for(auto &&i:n)
#define all(x) (x).begin(),(x).end()
#define SORT(x) sort(all(x))
#define REV(x) reverse(all(x))
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define LOWitr(x, n) lower_bound(all(x), n)
#define UPRitr(x, n) upper_bound(all(x), n)
#define cii(x) int x;cin >> x
#define cill(x) LL x;cin >> x
#define cis(x) string x;cin >> x
#define co(x) cout << x << endl
#define dump(x)  cout << #x << " = " << (x) << endl
#define truecheck assert
#define pb push_back
#define mp make_pair
#define F first
#define S second
using namespace std;
typedef long long LL;
typedef long long int64;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
template<class T,class U>bool chmax(T &a, const U &b) { if (a<b) { a=b; return true; } return false; }
template<class T,class U>bool chmin(T &a, const U &b) { if (b<a) { a=b; return true; } return false; }
template<class T,class U>void vcin(T &v, U n) { rep(i, n){cill(a);v.pb(a);} }
const int INF = 1e9, MOD = 1e9 + 7;
const LL LLINF = 1e16;

int main() {
    fastcin();
    cout << fixed << setprecision(10);
    cill(n);
    cill(m);
    VL vl;
    vcin(vl, n);
    function<int64(int64, int64)> gcd = [&gcd](int64 x, int64 y){
        if (x < y) swap(x, y);
        if (x % y == 0) return y;
        return gcd(y, x % y);
    };
    function<int64(int64, int64)> lcm = [&gcd](int64 x, int64 y){
        return x * y / gcd(x, y);
    };
    LL two = 0;
    LL f = vl[0];
    while (f % 2 == 0){
        f /= 2;
        two++;
    }
    LL base = pow(2, two);
    rep(i, n){
        if (vl[i] % base != 0 || vl[i] % base % 2 != 0){
            co(0);
            return 0;
        }
    }
    LL lc = 1;
    rep(i, n){
        lc = lcm(lc, vl[i]/2);
    }
    LL ans = (m / lc + 1) / 2;
    co(ans);
    return 0;
}