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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
typedef long long ll;
#define rp(i, f, t) for (int i = f; i < t; i++)
#define pr(i, f, t) for (int i = t - 1; i >= f; i--)
long long gcd(long long a, long long b)
{
    return b ? gcd(b, a % b) : a;
}
map<long long, long long> prime_factor(long long n)
{
    map<long long, long long> res;
    for (long long i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            ++res[i];
            n /= i;
        }
    }
    if (n != 1)
        res[n] = 1;
    return res;
}
int main()
{
    ll n, m;
    cin >> n >> m;
    vector<ll> a(n);
    rp(i, 0, n) cin >> a[i];
    ll g = a[0] / 2;
    rp(i, 1, n) g = gcd(g, a[i] / 2);
    ll l = (a[0] / 2) * (a[1] / 2) / gcd(a[0] / 2, a[1] / 2);
    rp(i, 2, n)
    {
        l = l * (a[i] / 2) / gcd(l, a[i] / 2);
    }
    // cout << l << endl;
    ll e = 0, o = 0;
    rp(i, 0, n)
    {
        if ((l / (a[i] / 2)) % 2 == 0)
            e++;
        else
            o++;
    }
    ll ans = 0;
    if (o == n && m >= l)
    {
        ans = (m - l) / (l * 2) + 1;
    }
    //   cout << ans << endl;
}