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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)x.size())
template<class T>bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,T b){if(a>b){a=b;return 1;}return 0;}

/*{
  }*/

using vi = vector<int>;
using vvi = vector<vi>;
using P = pair<int,int>;

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

signed main() {
  int n, m;
  cin >> n >> m;

  vi a(n);
  rep(i, 0, n){
    cin >> a[i];
  }
  
  vi b(n);
  rep(i, 0, n){
    int t = a[i];
    while(t%2 == 0){
      t /= 2;
      b[i]++;
    }
    
    if(i > 0 and b[i-1] != b[i]){
      cout << 0 << endl;
      return 0;
    }

    a[i] /= 2;
  }


  int lcm = a[0];
  rep(i, 1, n) lcm = lcm * a[i] / gcd(a[i], lcm);

  int ans = m / lcm / 2 + 1;
//  if(m < lcm) ans = 0;
  cout << ans << endl;

  return 0;
}