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)

/*
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*/
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define LONGMAX 1e18
#define ll long long
#define dd double
#define MOD 1000000007
#define nl cout<<endl
#define rt ll Q[100005]={0}
#define mp make_pair
#define endl "\n"
#define test() ull t;cin>>t;while(t--)
#define pb push_back
#define ff first	
#define ss second
#define all(v) v.begin(),v.end()
#define s(ar,n) sort(ar,ar+n)
#define rs(ar,n) sort(ar, ar+n, greater<ll>()) 
#define oa(a,n) for(ll i=0;i<n;i++)cout<<a[i]<<" ";nl
#define cn(a,n) for(ll i=0;i<n;i++)cin>>a[i];
#define ov(a,n) for(ll i=n-1;i>=0;i--)cout<<a[i]<<" ";nl
#define maxa(ar,N) *max_element(ar,ar+N)
#define mina(ar,N) *min_element(ar,ar+N)
#define INF 1e9
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
ll powe(ll x,ll y){ll res=1;while(y>0){if(y&1)res=res*x;y=y>>1;x=x*x;}return res;}
ll mpowe(ll n,ll m){if(m==0)return 1;ll t=powe(n,m/2);t%=MOD;if(m%2==0)return (t*t)%MOD;return (((t*t)%MOD)*n)%MOD;}
ll logtwo(ll n){if(n==1)return 0;return logtwo(n/2)+1;}
void ceilsearch(ll ar[],ll low,ll high,ll x,ll &y){ll mid=(low+high)/2;if(low>high)return;if(ar[mid]>=x){y=mid;ceilsearch(ar,low,mid-1,x,y);}else{ceilsearch(ar,mid+1,high,x,y);}}
int main()
{
	fastio();
	ll N,M;
	cin>>N>>M;
	ll ar[N];
	cn(ar,N);
	ll lcm=ar[0];
	for(ll i=1;i<N;i++)
	{
	    ll gc=__gcd(lcm,ar[i]);
	    lcm=(lcm*ar[i])/gc;
	    if(lcm>1e10)
	    {
	        cout<<0<<endl;
	        return 0;
	    }
	}
	ll k=lcm;
	lcm/=2;
	for(ll i=0;i<N;i++)
	{
	    if((lcm-ar[i]/2)%ar[i]!=0)
	    {
	        cout<<0<<endl;
	        return 0;
	    }
	}
	ll kj=M-lcm;
	kj=kj+(k-1);
	kj/=k;
	cout<<kj<<endl;
	cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}