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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

template<class T>
T gcd(T n,T m){
    if(n>m) return gcd(m,n);
    else if(n==0) return m;
    else return gcd(m-n*(m/n),n);
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    istream& in(cin);
    ostream& out(cout);
    int n,m;
    in>>n>>m;
    vector<ll> A(n);
    for(auto&i:A) in>>i;
    sort(A.rbegin(),A.rend());
    ll a,x=1,d;
    fr(i,n){
        a=A[i];
        if(x%a==0&&(x/a)%2==0)
            return cout<<0<<endl,0;
        d=gcd(a,x);
        x*=a;
        x/=d;
    }
    cout<<(m+x/2)/x<<endl;

}