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;
const int mod=998244353;
const int mex=2005;
#define ll long long 
#define test int t;cin>>t;while(t--)
#define fast  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define fo(i,a,n) for(int i=a;i<n;i++)
#define rfo(i,a,b) for(int i=a;i>=b;i--)
#define bg begin()
#define en end()
#define fi first
#define se second
#define ub upper_bound
#define lb lower_bound
#define pb push_back
#define veci vector<int>
#define veclli vector<long long int> 
#define all(x) x.begin(),x.end()
#define sci(x) scanf("%d",&x);
#define scc(x) scanf("%c",&x);
#define scs(x) scanf("%s",x);
#define debug(arr,n) for(int i=0;i<n;i++) printf("%d ",arr[i]);
#define sz(x) x.size()
#define loop(x) for(auto it=x.begin();it!= x.end();it++) 
int main()
{
   ll int n,m;
   cin>>n>>m;
   ll int a[n];
   fo(i,0,n)
   {
       cin>>a[i];
       a[i]=a[i]/2;
   }
   int fg=0;
   ll int lm=(a[0]*a[1])/__gcd(a[0],a[1]);
   fo(i,2,n)
   {
      if(lm>m) fg=1;
       lm=(lm*a[i])/__gcd(lm,a[i]);
   }
   ll int ans=0;
   if(lm%2&&fg==0)
   {
       ll int k=m/lm;
       ans=(k+1)/2;
       cout<<ans<<endl;
   }
   else cout<<"0"<<endl;
}