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 <bits/stdc++.h>

using namespace std;

#define int long long
#define endl '\n'
#define MOD 1000000007
#define maxn 100010

int lcm(int a, int b) {
    return (a * b) / __gcd(a, b);
}

void solve() {
    int n, m;
    cin >> n >> m;

    vector<int> v(n);

    int ans = 1;
    for (int i = 0; i < n; i++) {
        cin >> v[i];
        ans = lcm(ans, v[i] / 2);
        v[i] /= 2;
    }

    bool flag = true;
    for (int i = 0; i < n; i++) {
        if ((ans / v[i]) % 2 == 0)
            flag = false;
    }
    if (!flag)
        cout << 0 << endl;
    else
        cout << (m / ans + 1) / 2 << endl;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
//    cin >> t;

    while (t--)
        solve();
    return 0;
}