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 <boost/multiprecision/cpp_dec_float.hpp>
//#include <boost/multiprecision/cpp_int.hpp>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <algorithm>
#include <bitset>
#include <iomanip>
using namespace std;
//namespace mp = boost::multiprecision;
//using llll = mp::cpp_int;
// 仮数部が1024ビットの浮動小数点数型(TLEしたら小さくする)
//using Real = mp::number<mp::cpp_dec_float<1024>>;
using ll = long long;
const double EPS = 1e-10;
const ll MOD = 7+(1e+9);

int gcd(int a, int b)
{
    if (a < b) {
        a ^= b;
        b ^= a;
        a ^= b;
    }

    return b ? gcd(b, a % b) : a;
}

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

int main(){
  //cout<<fixed<<setprecision(10);
  ll n,m,ans=0;
  cin>>n>>m;
  vector<ll>v(n,0);
  for(ll i=0;i<n;i++){
    cin>>v[i];
    v[i]=v[i]/2;
  }
  ll t=v[0];
  for(ll i=1;i<n;i++){
    t=lcm(t,v[i]);
  }
  cout<<m/t/2+m/t%2<<endl;
}