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;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep(i, n) for(ll i = 0;i < n;i++)
#define all(i) i.begin(), i.end()
template<class T, class U> bool cmax(T& a, U b) { if (a<b) {a = b; return true;} else return false; }
template<class T, class U> bool cmin(T& a, U b) { if (a>b) {a = b; return true;} else return false; }

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

ll f(ll n) {
  ll cnt = 0;
  while (n%2 == 0) {
    n /= 2;
    cnt++;
  }
  return cnt;
}

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

  ll n, m;
  cin >> n >> m;
  vl a(n);
  rep(i, n) {
    cin >> a[i];
    a[i] /= 2;
  }

  rep(i, n) {
    if (f(a[0]) != f(a[i])) {
      cout << 0 << endl;
      return 0;
    }
  }

  ll t = 1;
  rep(i, n) {
    t = lcm(t, a[i]);
  }
  cout << m/t-m/(2*t) << endl;
}