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>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define all(a) a.begin(),a.end()
#define rall(c) (c).rbegin(),(c).rend()
#define mp make_pair
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef long double ld;
const ll inf=1e9+7;
const ll mod=998244353;
ll gcd(ll a,ll b){
    ll k=max(a,b),s=min(a,b);
    if(k%s==0)return s;
    return gcd(s,k%s);
}
int main(){
    ll n,m;cin>>n>>m;
    vector<ll>a(n);
    vector<ll>z(0);
    rep(i,n){
        cin>>a[i];
        a[i]/=2;
    }
    rep(i,n){
        ll k=a[i];
        ll cnt=0;
        while(k%2==0){
            cnt++;
            k/=2;
        }
        z.pb(cnt);
    }
    ll lcm=1;
    rep(i,n){
        ll k=lcm;
        lcm=(k/gcd(a[i],k)*a[i]);
        if(lcm>m){
            m=0;
            break;
        }
    }
    sort(all(z));
    if(z[0]!=z[n-1])m=0;
    cout<<(m/lcm+1)/2<<endl;
}