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 <stdio.h>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <numeric>
#include <complex>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <cassert>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
const double EPS = 1e-9;
typedef vector<int> vint;
typedef vector<vector<int>> v2int;
typedef vector<ll> vll;
typedef vector<vector<ll>> v2ll;
typedef list<int> liint;
typedef pair<int, int> pint;
const int INF = int(2e9);
const ll LINF = ll(2e9) * ll(2e9);
#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for(int i = x; i < n; i++)
template<class T, class C> void chmax(T& a, C b) { a > b ? : a = b; }
template<class T, class C> void chmin(T& a, C b) { a < b ? : a = b; }


ll gcd(ll a, ll b) {
    if (a < b) return gcd(b, a);
    ll r;
    while ((r = a%b)) {
        a = b;
        b = r;
    }
    return b;
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll N, M;
    cin >> N >> M;
    vll A(N);
    rep(i, N) cin >> A[i];

    ll L = A[0];
    REP(i, 1, N) {
        L = lcm(L, A[i]);
        if (M < L / 2) {
            cout << 0 << endl;
            return 0;
        }
    }
    rep(i, N) {
        if ((L - A[i]) % (2 * A[i]) != 0) {
            cout << 0 << endl;
            return 0;
        }
    }
    if (M < L / 2) {
        cout << 0 << endl;
        return 0;
    }
    ll ans = 1 + (M - (L / 2)) / L;
    cout << ans << endl;
    return 0;
}