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;
typedef long long ll;
#define int ll
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define ALL(a) (a).begin(), (a).end()
typedef vector<ll> vi;
typedef vector<vector<ll>> vvi;
typedef pair<ll, ll> pii;
const ll INF = 1LL << 58;
const ll MOD = 1000000007;
typedef vector<pii> vpii;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

// 最大公約数
int gcd(int a, int b)
{
    if (a < b)
        swap(a, b);
    if (a % b == 0)
        return (b);
    else
        return (gcd(b, a % b));
}

//最小公倍数
int lcm(int a, int b)
{
    return a * b / gcd(a, b);
}

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

    vi A(N);
    REP(i, N)
    {
        cin >> A.at(i);
        A.at(i) /= 2;
    }
    int llcm = lcm(A.at(0), A.at(1));
    REP(i, N)
    {
        llcm = lcm(llcm, A.at(i));
    }

    int ans = 0;
    int ii = 0;
    int tmp;

    while (true)
    {
        tmp = llcm * (2 * ii + 1);
        if (tmp <= M)
        {
            ans++;
        }
        else
        {
            break;
        }
        ii++;
    }
    cout << ans << endl;
    return 0;
}