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> // ver2.3.9
#include <complex.h>
//#include <boost/math/tools/minima.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
#define fmin brent_find_minima
#define int long long
#define endl "\n"
#define ALL(v) (v).begin(),(v).end()
#define COUNT(a,k) upper_bound(ALL(a),k)-lower_bound(ALL(a),k)
#define BIGGER(a,k) a.end()-upper_bound(ALL(a),k)
#define SMALLER(a,k) lower_bound(ALL(a),k)-a.begin()
#define Vi vector<int>
#define VVi vector<Vi>
#define Vs vector<string>
#define Pii pair<int,int>
#define VPii vector<Pii>
#define Tiii tuple<int,int,int>
#define VTiii vector<Tiii>
#define PQi priority_queue<int>
#define PQir priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define itos to_string
#define stoi stoll
#define FI first
#define SE second
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define cyes cout<<"yes"<<endl
#define cno cout<<"no"<<endl
#define sortr(v) sort(v,greater<>())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define leng(n) (int)(log10(n)+1)
#define dem(a,b) ((a+b-1)/(b))
#define Vin(a) rep(iI,0,a.size())cin>>a[iI]
#define INF 3000000000000000000  //  3.0*10^18(MAXの1/3くらい)
#define MAX LLONG_MAX
#define PI 3.141592653589793238462
#define MOD 1000000007  //  10^9 + 7

using namespace std;
/* fmin(f,L,R,100) で関数の最小値を(x,min)で返す　*/
//using boost::math::tools::brent_find_minima;
/* cpp_int (任意の長さのint) */
//using boost::multiprecision::cpp_int;

/* デバッグ用 */
void Vout(auto a){if(a.size()==0) cout<<"."<<endl;
else{rep(i,0,a.size()-1)cout<<a[i]<<' ';if(a.size()>0)cout<<a.back()<<endl;}return;}
void Verr(auto a){if(a.size()==0) cerr<<"."<<endl;
else{rep(i,0,a.size()-1)cerr<<a[i]<<' ';if(a.size()>0)cerr<<a.back()<<endl;}return;}
void VVout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())Vout(a[i]);}return;}
void VVerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())Verr(a[i]);}return;}
void VPout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())cout<<a[i].FI<<' '<<a[i].SE<<endl;}return;}
void VPerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())cerr<<a[i].FI<<' '<<a[i].SE<<endl;}return;}

/* 便利関数 */
int gcd(int a,int b){return b?gcd(b,a%b):a;} // 最大公約数gcd
int lcm(int a,int b){return a/gcd(a,b)*b;} // 最小公倍数lcm
int mypow(int x, int n, int m){ // 累乗x^n(mod m) O(log(n))
if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int scomb(int n, int r){if((n-r)<r)r=n-r; // nCr(小さい場合)
int a=1;for(int i=n;i>n-r;--i){a=a*i;}for(int i=1;i<r+1;++i){a=a/i;}return a;}
int comb(int n, int r){if((n-r)<r)r=n-r; // nCr(%MOD)
int a=1;for(int i=n;i>n-r;--i){a=a*i%MOD;}for(int i=1;
i<r+1;++i){a=a*mypow(i,MOD-2,MOD)%MOD;}return a%MOD;}
Vi stpowV(){Vi a(100001);
a[0]=1;repeq(i,1,100000)a[i]=a[i-1]*i%MOD;return a;}
//Vi stpow = stpowV();  // 階乗配列(%MOD)
Vi Vsum(Vi &v){Vi S(v.size()+1);rep(i,1,S.size())S[i]+=v[i-1]+S[i-1];return S;}// 累積和
int digit(int k,int i){string s = itos(k);return s[s.size()-i]-'0';} // i桁目の数字
void press(auto &v){v.erase(unique(ALL(v)),v.end());} // 圧縮
Vi zip(Vi b){int Z=b.size(); // 座標圧縮
Pii p[Z+10];int a=b.size();Vi l(a);for(int i=0;i<a;i++)
p[i]=mp(b[i],i);sort(p,p+a);int w=0;for(int i=0;i<a;i++)
{if(i&&p[i].first!=p[i-1].first)w++;l[p[i].second]=w;}return l;}
Vi beki2V(){Vi a(61);int q = 1;rep(i,0,61){a[i]=q;q*=2;}return a;}
//Vi beki2 = beki2V(); // 2^n配列
int modiv(int a,int b){return a*mypow(b,MOD-2,MOD)%MOD;} // a÷b(MOD環境)



signed main() {
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(12);
  int n,k;
  cin >> n >> k;
  Vi a(n);
  Vin(a);
  
  Vi b(n);
  rep(i,0,n) b[i] = a[i]/2;
  
  int all = a[0];
  int han = a[0]/2;
  
  rep(i,0,n) {
    
    if(i!=0){
      if((all/2)%a[i]==0) {
        cout << 0 << endl;
        return 0;
      }
    }
    
    all = lcm(all,a[i]);
    han = lcm(han,a[i]/2);
  }
  
  int ans;
  
  ans = k/han;
  ans -= k/all;
  
  int am = b[0]%2;
  rep(i,0,n){
    if(b[i]%2 != am){
      cout << 0 << endl;
      return 0;
    }
  }
  
  
  cout << ans << endl;
  
  return 0;
}