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: Runtime Error)

#define _GLIBCXX_DEBUG
#include <iostream>
#include <vector>
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 f(int x) {
  int res = 0;
  while (x % 2 == 0) {
    x /= 2;
    res++;
  }
  return res;
}

int main() {
  int n, m;
  cin >> n >> m;
  vector<ll> a(n + 1);
  REP(i, n) cin >> a[i];

  // a -> a'にしている
  REP(i, n) {
    if (a[i] % 2 == 1) {
      cout << 0 << endl;
      return 0;
    }
    a[i] /= 2;
  }

  /* a' -> a''、つまり上記で言う所の2^tで割ってる
     これをやると一々a'の最大に最小公倍数を
     合わせるとか考えることがなくなるので楽
  */
  int t = f(a[0]);
  REP(i, n) {
    a[i] >>= t;  // a[i] /= 2^t
  }
  m >>= t;

  ll l = 1;
  REP(i, n) {
    l = lcm(l, a[i]);
    if (l > m) {
      cout << 0 << endl;
      return 0;
    }
  }
  // m -> m'
  m /= l;

  /*
    これはm'以下の奇数の数、つまり
    mを最小公倍数で割った時の奇数倍を表す
    (奇数倍の数=倍数中の奇数の数を求めれば良い)
  */
  ll ans = (m + 1) / 2;
  cout << ans << endl;

  return 0;
}