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>
#include <boost/multiprecision/cpp_int.hpp>
#define rep(X,N) for(ll X = 0; X < (N); X++)
#define PI (acos(-1.0))
#define pback push_back
#define mpair make_pair
#define MODN 1000000007
#define ALL(V) (V).begin(),(V).end()

typedef long long ll;

using namespace std;
using namespace boost::multiprecision;

ll gcd(ll a, ll b){
    if(a < b) return gcd(b, a);

    if(a % b == 0){
        return b;
    }else{
        return gcd(b, a % b);
    }
}

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

int main(){

    int n,m;
    cin >> n >> m;
    
    vector<int> a;

    int maxa = 0;

    ll lcmi = 1;

    rep(i,n){
        int tmp;
        cin >> tmp;

        lcmi = lcm(lcmi, tmp / 2);

        maxa = max(maxa, tmp / 2);

        if(lcmi > (ll)m){
            break;
        }

        a.pback(tmp / 2);
    }

    bool oddeven = true;

    

    rep(i,n){
        
        if(a[i] == 0 || (lcmi / a[i]) % 2 == 0){
            oddeven = false;
            break;
        }
    }

    int ans = 0;

    cerr << lcmi << endl;

    if(oddeven){

        for(int i = lcmi; i <= m; i = i + lcmi * 2){
            ans++;
        }

    }

    cout << ans << endl;

    return 0;
}