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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define P pair<ll, ll>
using ll = int64_t;
using namespace std;
#define ketasuu(n) fixed << setprecision(n)
#define btoe(p) p.begin(), p.end()
#define etob(p) p.rbegin(), p.rend()

//define gcd and lcm (using 64bit)
ll gcd(ll a, ll b)
{
    if (a < b)
    {
        //^はxor(bitwise operator)。以下三行でswap(a,b)を行っている。
        a ^= b;
        b ^= a;
        a ^= b;
    }
    //only when b==0,this program finds that b means false and returns a.
    return b ? gcd(b, a % b) : a;
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}

int main(){
    ll n, m;
    cin >> n >> m;
    bool ok=true;
    vector<ll> p(n);
    ll l=1;
    rep(i,n){
        cin>>p[i];
        p[i]/=2;
        l=lcm(l,p[i]);
        if(l>m){
            ok=false;
            break;
        }
    }
    rep(i,n){
        if(((l/p[i])%2)==0){
            ok=false;
            break;
        }
    }
    if(!ok){
        cout<<0<<endl;
        return 0;
    }
    ll ans=(m+l)/(2*l);
    cout<<ans<<endl;
    return 0;
}