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;
using ll = long long;
const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
const ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
ll myctoi(char C) {return C - '0';}
char myitoc(ll N) {return '0' + N;}
#define FINALANS(A) {cout << (A) << '\n'; exit(0);}

ll mygcd(ll A, ll B)
{
  ll a = A, b = B;
  while (a > 0 && b > 0)
  {
    b %= a;
    swap(a, b);
  }
  return b;
}

ll mylcm(ll A, ll B)
{
  ll G = mygcd(A, B);
  return A / G * B;
}

int main()
{
  ll N, M;
  cin >> N >> M;
  vector<ll> A(N);
  for (ll i = 0; i < N; i++)
  {
    cin >> A.at(i);
  }

  ll L = A.front() / 2;
  for (ll i = 1; i < N; i++)
  {
    L = mylcm(L, A.at(i) / 2);
  }

  //cerr << L << endl;

  if (L % 2 == 0)
    FINALANS(0);

  ll ans = divceil(M / L, 2);
  cout << ans << endl;
}