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>
using namespace std;
#define repd(i,a,b) for (ll i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef long long ll;
const long long INF = 1LL << 60;
typedef pair<int, int> P;

ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

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

int main()
{
    ll N, M;
    cin >> N >> M;

    vector<ll> a(N);
    rep(i, N) cin >> a[i];

    ll tmp = lcm(a[0], a[1]);
    repd(i, 2, N) {
        tmp = lcm(tmp, a[i]);
    }


    ll cp = tmp / 2;
    ll ans = 0;
    if (cp <= M) {
        ans++;
        M -= cp;
        if (tmp <= M) ans += (ll)(M / tmp);
    }

    sort(all(a));
    rep(i, N - 1) {
        ll cmp = a[i] * 2LL;
        if (cmp == a[i + 1LL]) {
            ans = 0;
            break;
        }
        if (a[N - 1] % 2 != a[i] % 2) {
            ans = 0;
            break;
        }
    }
    cout << ans << endl;
    return 0;
}