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)

//author:luckYrat(twitter:@luckYrat_)
//#include <bits/stdc++.h>

//def
#include <iostream>

#include <cmath>
#include <algorithm>
#include <iomanip>

//array
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <utility>
#include <climits>



using namespace std;

typedef pair<int,int> P;
typedef long long ll;

#define anyfill(n,s) setw(n) << setfill(s)
#define loop(s) for(int i = 0; s > i; i++)
#define rep(i,q) for(int i = 0; (q) > i; i++)
#define repp(i,n,q) for(int i = n; (q) > i; i++)
#define dep(i,q) for(int i = (q); 0 < i; i--)


#define MAX 1000000000
#define MOD 1000000007
#define EPS (1e-10)

#define pb push_back
#define fir first
#define scn second
#define ednl endl

#define YesNo(a) (a?"Yes":"No")
#define YESNO(a) (a?"YES":"NO")
#define yesno(a) (a?"yes":"no")



/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる
vector.unique()は隣接した同じ要素を削除するので先にソートをする

計算量は変わらないが楽できるシリーズ
C++11以降でmin({a,b,c...})で複数個のmic,maxを処理できる
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる

*/
template <typename T>
T gcd(T a,T b){
  if(a%b==0)return b;
  else return gcd(b,a%b);
}
template <typename T>
T lcm(T a,T b){
  return a/gcd(a,b)*b;
}
__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
int main() { 
  int n,m;cin>>n>>m;
  int a[n];
  int lc = 1;
  for(int i = 0; n > i; i++){
    cin>>a[i];a[i]/=2;
    lc = lcm(lc,a[i]);
  }

  cout << ((m/lc)+1)/2 << endl;
}