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: Accepted)

#include<bits/stdc++.h>
using namespace std;

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

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

int main() {
  long long N,M;
  cin >> N >> M;
  vector<long long> a(N);
  
  for (int i=0; i<N; i++) {
    cin >> a[i];
    a[i] /= 2;
  }
  
  long long BLCM = a[0];
  for (int i=1; i<N; i++) {
    BLCM = lcm(BLCM,a[i]);
    if (BLCM > M) {
      cout << 0 << endl;
      return 0;
    }
  }
  
  long long val;
  for (int i=0; i<N; i++) {
    val = BLCM / a[i];
    if (val % 2 == 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  
  long long ans = M / BLCM;
  ans = (ans + 1) / 2;
    
  
  cout << ans << endl;
}