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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

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

int main() {
  // X = Ak/2 * (2*p + 1)
  // それぞれのAkが2で割れる回数が同じでなければならない
  int N, M;
  cin >> N >> M;
  vector<int> A(N);
  rep(i, N) cin >> A[i];
  int time = 0; // 2で割れる回数
  while (A[0] % 2 == 0) {
    time += 1;
    rep(i, N) {
      if (A[i] & 1) {
        cout << 0 << '\n';
        return 0;
      } else {
        A[i] /= 2;
      }
    }
  }
  if (time == 0) {
    cout << 0 << '\n';
    return 0;
  }
  ll l = 1;
  rep(i, N) {
    l = lcm(l, (ll)A[i]);
  }
  ll coef = 1;
  rep(i, time-1) coef *= 2;
  ll ans = M / l / coef- (M / 2 / l / coef);
  cout << ans << '\n';
  return 0;
}