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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using Vi = vector<int>;
using Vl = vector<ll>;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;

constexpr int I_INF = numeric_limits<int>::max();
constexpr ll L_INF = numeric_limits<ll>::max();

//==================================

ll mpow(ll a, ll n) {
    ll res = 1;
    for (; n > 0; n >>= 1) {
        if (n & 1) res *= a;
        a *= a;
    }
    return res;
}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N, M;
    cin >> N >> M;
    Vl A(N);
    for (int i = 0; i < N; i++) {
        int a;
        cin >> a;
        A[i] = a / 2;
    }

    unordered_map<ll, ll> mp;
    for (int i = 0; i < N; i++) {
        ll tmp = A[i];
        for (int j = 2; j * j <= tmp; j++) {
            if (tmp % j == 0) {
                ll cnt = 0;
                while (tmp % j == 0) {
                    tmp /= j;
                    ++cnt;
                }
                mp[j] = max(mp[j], cnt);
            }
        }
        if (tmp > 1) mp[tmp] = max(mp[tmp], 1LL);
    }

    ll l = 1;
    for (const auto& p : mp) {
        l *= mpow(p.first, p.second);
    }

    bool ok = true;
    for (int i = 0; i < N; i++) {
        if (l / A[i] % 2 == 0) {
            ok = false;
            break;
        }
    }

    if (!ok) {
        cout << 0 << endl;
        return 0;
    }

    cout << M / l - M / l / 2 << endl;

    return 0;
}