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>

#define ll long long int
#define ld long double
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define INF 0x3f3f3f3f
#define show(x) cerr<<#x<<" : "<<x<<'\n';
#define yogeshk972() cerr<<"\nTime Taken : "<<(float)(clock()-time_p)/CLOCKS_PER_SEC<<"\n";
clock_t time_p=clock();

using namespace std;
const int N= 1e3+10, mod= 1e9+7;
#define int ll

int gcdex(int a,int b,int& x,int &y){
  if(a==0){
    x=0; y=1;
    return b;
  }
  int x1,y1;
  int g= gcdex(b%a,a,x1,y1);
  x=y1- (b/a)*x1;
  y=x1;
  return g;
}

signed main() {
  ios_base::sync_with_stdio(false);cin.tie(NULL);

/*  #ifndef ONLINE_JUDGE
  freopen("in.txt","r",stdin);
  freopen("out.txt","w",stdout);
  #endif
*/  
  int n,m;
  cin>>n>>m;
  set<int> st;
  for(int i=0;i<n;i++){
    int  x; cin>>x;
    st.insert(x);
  }

  pair<int,int> p={ *begin(st)/2,*begin(st) }; 
  st.erase(st.begin());

  for(auto a:st){
    int x,y;
    int g=gcdex(p.yy,a,x,y);
    int c=a/2-p.xx;
    if( c%g!=0 ){
      cout<<"0";
      return 0;
    }
    c/=g;
    x*=c; y*=c;
    int temp= p.xx+p.yy*1ll*x;
    p.xx=temp; p.yy=(p.yy*1ll*a)/g;
   // cout<<p.xx<<" "<<p.yy<<"\n";
  }

  cout<<(m-p.xx)/p.yy + (p.xx<=m);

  return 0;
}