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;

#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
// pll pp; -> (pp.first,pp.second)
// vector<vl> pp(n,vl(2)); sort(pp.begin(),pp.end(),[](vl a,vl b){return a[0]<b[0];});

#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define RREP(i,n) for(ll (i)=(n)-1;(i)>=0;(i)--)
#define FOR(i,a,b) for(ll (i)=(a);(i)<(b);(i)++)
#define FORR(i,a,b) for(ll (i)=(b)-1;(i)>=(a);(i)--)
#define DEBUG(x) cout << #x << ": " << x << endl
#define CHMAX(a,b) a=max((a),(b))
#define CHMIN(a,b) a=min((a),(b))
/*}}}*/

ll gcd(ll a,ll b){
  if(a<b) swap(a,b);
  if(a%b==0) return b;
  return gcd(b,a%b);
}
int main() {
  ll n,m;
  cin >> n >> m;

  vl a(n);
  REP(i,n){
    ll ai;
    cin >> ai;
    a[i] = ai/2;
  }
  if(n==1){
    ll lcm = a[0];
    ll tmp = m/lcm;
    ll ans = tmp/2 + tmp%2;
    cout << ans << endl;
    return 0;
  }
  ll gd = gcd(a[0],a[1]);

  ll lcm = a[0]*a[1]/gd;
  if(lcm>m){
    cout << 0 << endl;
    return 0;
  }
  FOR(i,2,n){
    gd = gcd(lcm,a[i]);
    lcm *= a[i]/gd;
    if(lcm>m){
      cout << 0 << endl;
      return 0;
    }
  }

  ll tmp = m/lcm;
  ll ans;
  if(tmp%2==0) ans = tmp/2;
  else ans = tmp/2 + 1;
  
  //cout << fixed << setprecision(10);
  cout << ans << endl;
  return 0;
}