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 rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;

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

int main()
{
    long long N, M;
    cin >> N >> M;
    long long n = 1;
    vector<long long> a(N), pow2Count(N, 0);
    bool is_ok = true;
    rep(i, N)
    {
        cin >> a[i];
        int b = a[i] / 2;
        while (a[i] % 2 == 0)
        {
            a[i] /= 2;
            pow2Count[i]++;
        }
        n = lcm(n, b);
        if (n > M)
        {
            is_ok = false;
        }
    }

    rep(i, N)
    {
        if (pow2Count[0] != pow2Count[i])
            is_ok = false;
    }
    if (!is_ok)
    {
        cout << 0 << endl;
        return 0;
    }
    long long ans = max((M + n - 1) / (2 * n), (ll)0);
    if (M == n)
        ans++;
    cout << ans << endl;
    return 0;
}