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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>

using namespace std;

#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)

using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

ll gcd(ll a, ll b) {
    ll d = 1;
    if (b != 0) {
        d = gcd(b, a % b);
    } else {
        return a;
    }
    return d;
}
ll lcm(int a, int b) {
    ll tmp = a * b;
    return tmp / gcd(a, b);
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n >> m;
    vector<int> a;
    for (int i = 0; i < n; i++)
    {
        ll tmp;
        cin >> tmp;
        a.push_back(tmp);
    }
    
    ll whole_lcm = a[0];
    for (int i = 1; i < n; i++)
    {
        whole_lcm = lcm(whole_lcm, a[i]);
    }
    
    if (whole_lcm % 2 == 0) {
        ll half_lcm = whole_lcm / 2;
        cout << (ll)m / half_lcm - (ll)m / whole_lcm << "\n";
    } else {
        cout << 0 <<"\n";
    }


}