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;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N,M; cin>>N>>M;
  vector<long long> A(N); for (auto& a : A) cin>>a;
  auto gcd = [](long long a, long long b) {
    if (a > b) swap(a,b);
    long long r;
    while (r = b%a) {
      b = a;
      a = r;
    }
    return a;
  };
  auto lcm = [&](long long a, long long b) {
    return (a*b)/gcd(a,b);
  };
  auto solve = [&]() -> long long {
    auto lcmab = A[0];
    for (int i = 1; i < N; i++) {
      lcmab = lcm(lcmab, A[i]);
      if (lcmab > M*2)
        return 0;
    }
    auto d = lcmab/2;
    auto k = M/d;
    return (k+1)/2;
  };
  cout << solve() << endl;

  return 0;
}