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 rep(i,r) for(int i = 0; i < (r); i++)
#define pb push_back
#define ALL(a) a.begin(), a.end()
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define endl '\n'
using ll = long long;
using ull = unsigned long long;
const int maxn = 1e5 + 10;
ll A[maxn];

ll LCM(ll a, ll b) {
  return a / __gcd(a,b) * b;
}

int main() {
  fastio;
  ll N, M;
  cin >> N >> M;
  ll lcm = 1;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A[i] /= 2;
    lcm = LCM(lcm, A[i]);
    if (lcm > M) {
      cout << 0 << endl;
      return 0;
    }
  }

  // check to make sure ai/2 is a multiple of an odd
  for (int i = 0; i < N; i++) {
    if ((lcm / A[i]) % 2 == 0) {
      cout << 0 << endl; return 0;
    }
  }
  lcm = 1;
  rep(i, N) {
    lcm = LCM(lcm, A[i]*2);
  }
  cerr << "LCM: " << lcm << endl;
  ll res = M / lcm;
  // res = (res + 2 - 1) / 2;
  cout << res << endl;
  
}