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;

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

long long int lcm(long long int a, long long int b)
{
  long long int g = gcd(a, b);
  return a / g * b;
}

int main()
{
  long long int N, M;
  cin >> N >> M;
  vector<long long int> a(N);
  for (int i = 0; i < N; i++)
  {
    long long int b;
    cin >> b;
    b /= 2;
    a[i] = b;
  }
  bool can = true;
  long long int num = 0;
  long long int div = 2;
  while (a[0] % div == 0) {
    num++;
    div *= 2;
  }
  for (int i = 1; i < N; i++)
  {
    long long int tmp = 0;
    div = 2;
    while (a[i] % div == 0) {
      tmp++;
      div *= 2;
    }
    if (tmp != num)
    {
      can = false;
      break;
    }
  }
  if (!can)
    cout << 0 << endl;
  else
  {
    long long int T = a[0];
    for (int i = 1; i < N; i++)
    {
      T = lcm(a[i], a[i - 1]);
    }
    long long int ans = (M / T + 1) / 2;
    cout << ans << endl;
  }
}