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 <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <bitset>
#include <complex>
#include <functional>

// output
#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
// utility
#define ALL(i) (i).begin(), (i).end()
#define FOR(i, a, n) for(int i=(a);i<(n);++i)
#define RFOR(i, a, n) for(int i=(n)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define IN(a, x, b) (a<=x && x<b)
#define OUT(a, x, b) (x<a || b<=x)
template<class T> inline T chmax(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T chmin(T& a, const T b) { return a = (a > b) ? b : a; }

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const int MOD = 1000000007;
/* const int MOD = 998244353; */
const int INF = 1e18;
const double PI = acos(-1);

using namespace std;

int lcm(int x, int y){
  return x/__gcd(x, y)*y;
}

signed main() {
  int N, M; 
  cin >> N >> M;

  vector<int> A(N);
  REP(i, N) cin >> A[i];

  int l = 1;
  REP(i, N) l = lcm(l, A[i]/2);

  cout << M/l-M/(l*2) << endl;

  return 0;
}